C++ Binary Search Tree implementation does not add every element - c++

I have a simple C++ implementation of a BST. I am only trying to add numbers and to print them out, in order. The problem is that out of the 16 numbers I try to add, I am only able to add 12 (leaving out 32, 15, 14 and 3). The output from my console is shown below:
Printing the tree before adding numbers: The list is
empty The key 32 has already been added. The key 15 has already
been added. The key 14 has already been added. The key 3 has
already been added. Printing the tree in order after adding
numbers: 2 4 21 50 52 64 70 76 80 83 87 100 Program ended with
exit code: 0
#include <iostream>
using namespace std;
class BST {
private:
struct node {
int data;
node * left;
node * right;
};
node * root;
void addLeafPrivate(int data, node * n);
void printInOrderPrivate(node *);
public:
BST();
node * createLeaf(int data);
void addLeaf(int data);
void printInOrder();
};
int main() {
int TreeKeys[16]= {50, 76, 21, 4, 32, 64, 15, 52, 14, 100, 83, 2, 3, 70, 87, 80};
BST bst;
cout << "Printing the tree before adding numbers: \n";
bst.printInOrder();
for (int i = 0; i < 16; i++) {
bst.addLeaf(TreeKeys[i]);
}
cout << "Printing the tree in order after adding numbers: \n";
bst.printInOrder();
return 0;
}
BST::BST() {root = NULL;}
BST::node * BST::createLeaf(int data) {
node * n = new node;
n->data = data;
n->right = NULL;
n->left = NULL;
return n;
}
void BST::addLeaf(int data) {
addLeafPrivate(data, root);
}
void BST::addLeafPrivate(int data, node * n) {
if (root == NULL) {
root = createLeaf(data);
}
else if (data < n->data) { // add recursively on left left side.
if (n->left != NULL){
addLeafPrivate(data, n->left);
}
else { // if left is empty
n->left = createLeaf(data);
}
}
else if (data > root->data) { // add recursively on right left side.
if (n->right != NULL) {
addLeafPrivate(data, n->right);
}
else { // right is empty
n->right = createLeaf(data);
}
}
else {
cout << "The key " << data << " has already been added.\n";
}
}
void BST::printInOrder() {
printInOrderPrivate(root);
}
void BST::printInOrderPrivate(node * n) {
if (n != NULL) {
if (n->left != NULL) {
printInOrderPrivate(n->left);
}
cout << n->data << " ";
if (n->right != NULL) {
printInOrderPrivate(n->right);
}
}
else {
cout << "The list is empty\n";
}
}

else if (data > root->data) { // add recursively on right left side.
should be
else if (data > n->data) { // add recursively on right left side.
The problems start when you come to 32. Since 21 is already left of 50, your algorithm thinks it's already inserted 32, when you do root->data instead of the correct n->data, instead of comparing the data values and throwing an exception.
So: Check right and left for null, compare if data is greater or less AND check for equality. Doing so lets you more easily find bugs like this.

Related

Delete a Node from C++ Binary Search Tree (class not struct)

I'm trying to manage a BST in C++ for my academic purpose.
I'm not having problems anywhere except for the DeleteNode function, also
I chose to implement this data structure with a class and not with a struct.
The problem is, that I cant figure it out how to make properly work the delete function, often I got 0xDDDDDDDDD error my debugger say, sometimes I can delete the node, sometimes my program crash.
I think that's a possible problem of pointer, but I just can't figure it out where I'm doing wrong.
Here's my delete node function, the one I'm getting serious trouble about:
EDIT: The no-son delete case works perfectly, the one i'm getting mad about is the one-son-case delete.
//function that delete a selected node
void DeleteNode(TreeNode* root,int key) {
/*we got three case here:*/
//until we find the right node with value in the tree
if (root->getValue() != key && root != nullptr) {
if (root->getValue() > key) {
DeleteNode(root->Left, key);
}
else if (root->getValue() < key) {
DeleteNode(root->Right, key);
}
}
else { //when we found the right node, then operate
/* THIS WORKS PERFECTLY! */
//first case: our node got no right and left son
if (!root->Left && !root->Right) {
TreeNode* tmp = root->Father;
if (tmp->Left == root) { //if the son is a left son
tmp->Left = nullptr;
delete root;
}
else if (tmp->Right == root) { //if the son is a right son
tmp->Right = nullptr;
delete root;
}
}
//second case: our node got a left but no right son
/* THIS ONE DOESN'T WORK. */
else if (!root->Right) {
TreeNode *tmp = root;
root = root->Left; //new root is the left son of the root
root->Father = tmp->Father; //linking the father to the new son
tmp->Father->Left = root; //linking the son to the new father
delete tmp;
std::cout << "Erased!" << std::endl;
}
else if (!root->Left) {
TreeNode *tmp = root;
root = root->Right; //new root is the right son of the root
root->Father = tmp->Father; //linking the father to the new son
tmp->Father->Right = root; //linking the son to the new father
delete tmp;
std::cout << "Erased!" << std::endl;
}
}
}
I tried a lot of combination, but the result are the same every time: it crashes on the first occurrence of the InOrder display function. (and when it does not, the function just delete the first nodes and then crash when i try to delete a new one.)
Here's a simple main where i'm trying to act the delete:
int main()
{
TreeNode root;
root.insertNode(&root,50);
root.insertNode(&root,30);
root.insertNode(&root,20);
root.insertNode(&root,40);
root.insertNode(&root,70);
root.insertNode(&root,60);
root.insertNode(&root,80);
for (int i = 0; i < 5; i++) {
int n;
cin >> n;
root.DeleteNode(&root, n);
cout << "In-Order: "; root.inOrder(&root);
cout << endl;
cout << "Pre-Order: "; root.preOrder(&root);
cout << endl;
cout << "Post-Order: "; root.postOrder(&root);
cout << endl;
}
}
Here's my full BST code (except the delete one that i submitted before, just for being more complete in the understanding of my code)
class TreeNode {
private:
int value;
TreeNode* Left;
TreeNode* Right;
TreeNode* Father;
public:
//constructor
TreeNode() {
this->Right = nullptr;
this->Left = nullptr;
this->Father = nullptr;
}
TreeNode(int value) {
this->value = value;
this->Right = nullptr;
this->Left = nullptr;
this->Father = nullptr;
}
//functions
int getValue() { return value; }
void setValue(int value) { this->value = value; }
//function to create a new node and insert a value into it
TreeNode* insertNode(TreeNode* root, int value) {
if (root->getValue() == NULL) {
root->setValue(value);
root->Father = nullptr;
}
else {
if (value > root->getValue()) {
if (root->Right) {
insertNode(root->Right, value);
}
else
root->Right = new TreeNode(value);
root->Right->Father = root;
}
else if (value < root->getValue()) {
if (root->Left) {
insertNode(root->Left, value);
}
else
root->Left = new TreeNode(value);
root->Left->Father = root;
}
}
return root;
}
//function to search a value into a BST
TreeNode* SearchNode(TreeNode* root, int key) {
if (root->getValue() == key) {
return root;
}
else if (root->getValue() < key) {
if (root->Right) {
SearchNode(root->Right, key);
}
else return nullptr;
}
else if (root->getValue() > key) {
if (root->Left) {
SearchNode(root->Left, key);
}
else return nullptr;
}
}
//function that return the height of the tree
int TreeHeigth(TreeNode* root) {
int heigth;
if (root == nullptr) {
return 0;
}
else {
return heigth = 1 + max(TreeHeigth(root->Left), TreeHeigth(root->Right));
}
}
//function that returns the number of the nodes
int CountTreeNode(TreeNode* root) {
if (root == nullptr) {
return 0;
}
else {
return CountTreeNode(root->Left) + CountTreeNode(root->Right) + 1;
}
}
//function that returns the minimum values into the tree
TreeNode* MinimumNode(TreeNode* root) {
if (root == nullptr) {
return nullptr;
}
while (root->Left != nullptr) {
root = root->Left;
}
return root;
}
//function that returns the maximum value into the tree
TreeNode* MaximumNode(TreeNode* root) {
if (root == nullptr) {
return nullptr;
}
while (root->Right != nullptr) {
root = root->Right;
}
return root;
}
//function that returns a successor of a given nodeb
TreeNode* SuccessorNode(TreeNode* node) {
//first case: our node got a rigth subtree:
if (node->Right != nullptr) {
return MinimumNode(node->Right);
}
//second case: our node doesnt got a right subtree: lets get
//upper in the tree until our node isn't a left child.
TreeNode* Ancestor = node->Father;
while (Ancestor != nullptr && node == Ancestor->Right) {
node = Ancestor;
Ancestor = Ancestor->Father;
}
}
//function tht returns a predecessor of a given node
TreeNode* PredecessorNode(TreeNode* node) {
//first case: (inverse to successor) our node got a left subtree:
if (node->Left != nullptr) {
return MaximumNode(node->Left);
}
TreeNode* Ancestor;
if (node->Father == nullptr)
return nullptr;
else
Ancestor = node->Father;
while (Ancestor != nullptr && node == Ancestor->Left) {
node = Ancestor;
Ancestor = Ancestor->Father;
}
return Ancestor;
}
//function that prints information about nodes
void InfoNode(TreeNode *root) {
root != nullptr ? std::cout << "Nodo corrente: " << root->getValue() << std::endl
: std::cout << "Nodo corrente: " << "NULL" << std::endl;
root->Father != nullptr? std::cout << "Padre: " << root->Father->getValue() << std::endl
: std::cout << "Padre: " << "NULL" << std::endl;
root->Left != nullptr ? std::cout << "Figlio SX: " << root->Left->getValue() << std::endl
: std::cout << "Figlio SX: " << "NULL" << std::endl;
root->Right!= nullptr ? std::cout << "Figlio DX: " << (root->Right)->getValue() << std::endl
: std::cout << "Figlio DX: " << "NULL" << std::endl;
}
//visits of a tree
void preOrder(TreeNode* root) {
if (root != nullptr) {
std::cout << root->getValue() << " ";
preOrder(root->Left);
preOrder(root->Right);
}
}
void inOrder(TreeNode* root) {
if (root != nullptr) {
inOrder(root->Left);
std::cout << root->getValue() << " ";
inOrder(root->Right);
}
}
void postOrder(TreeNode *root) {
if (root != nullptr) {
postOrder(root->Left);
postOrder(root->Right);
std::cout << root->getValue() << " ";
}
}
//max between 2 numbers
int max(int a, int b) {
return a > b ? a : b;
}
};
And there's the representation of the tree I'm trying to work on:
50
/ \
30 70
/ \ / \
20 40 60 80
Where i'm doing it wrong?
Look at this condition: root->getValue() != key && root != nullptr, This first calls getValue and after that checks root has legal value. swap them(root != nullptr && root->getValue() != key).
Finally I think you must change last line to tmp->Father->Left = root; to fix crash problem.
TreeNode *tmp = root;
root = root->Right; //new root is the right son of the root
root->Father = tmp->Father; //linking the father to the new son
tmp->Father->Right = root; //linking the son to the new father
PS: Also do this exchange for other side...
Note: This is true until root is left child of his father otherwise your code is true. Precisely you must check if root is left child if his father do tmp->Father->Left = root; else tmp->Father->Right = root;
Note: As you mentioned your code does not handle deletion of a node with two childern.
As there is already an answer giving you directions to correct the specific errors, I will try to focus on a suggestion that will help you avoid similar error all together:
Try separating your current function into two:
One that searches a node with specific key, for example: Node* search(int key) function that returns either a pointer to the node with wanted key or nullptr, or use the one that your already have.
One that deletes (and re-wires) the node passed as pointer and returns: next, previous, etc: Node* delete(Node* n).
Then call search, test against nulltpr, and if different, pass the returned pointer as an input argument in delete.
In this way you could easily detect on which phase is you problem: searching or deleting.
P.S.: figuring out re-wiring bugs is usually done through diagrams (boxes and arrows). Decide what you should do, separate it into steps and implement it.
Well, once one know that DEBUG version use sentinel value, it become much more trivial to find problems in code.
When and why will an OS initialise memory to 0xCD, 0xDD, etc. on malloc/free/new/delete?
0xDD is for dead memory. That is memory that has been already deleted. So when the debugger stop and it tells you that you have a bad pointer and the data contains a lot of 0xDD, you know the data has already been deleted. At that point, you should inspect class that contain the data to see if they are deleted too so you know which objects were deleted when the are embedded one inside another.
Be aware that sometime you might have some data that has been changed in part of the class if some operations use delete memory. Looking at memory pattern also help finding uninitialized memory and other similar problem.
Some other links:
Why does the not allocated memory is marked like 0xCC?
https://msdn.microsoft.com/en-us/library/974tc9t1.aspx
https://www.codeguru.com/cpp/w-p/win32/tutorials/article.php/c9535/Inside-CRT-Debug-Heap-Management.htm
http://www.nobugs.org/developer/win32/debug_crt_heap.html
In a case like yours, if you follow the good practice of writing unit tests, then it would even be more trivial to find the problem. In fact, if you do proper testing, then you will test all possible cases so you will know which cases fail and it would help you find where you might do something wrong.
I would like to add something to the answer of #Bonje Fir.
Sure it is a correct answer, but partially: I'll explain why.
He suggested to swap the last piece of code i wrote:
Case: we're in the right subtree, and we would like to erase 70 (because we don't have anymore the leaf node 60):
50
/ \
30 70
/ \ \
20 40 80
now, with the code that #Bonje Fir suggested us, we would got a problem here:
TreeNode *tmp = root;
root = root->Right; //new root is the right son of the root
root->Father = tmp->Father; //linking the father to the new son
tmp->Father->Left (instead of Right) = root; //linking the son to the new father
Because the code is saying, that once you updated the new root with his son, link the father of the previous root (that we saved in a tmp variable) with his left son. Then we would assist to something like this:
50
/ x
80 80
/ \
20 40
and that's inconsistent.
Now take a look on the other side, with the same code and without leaf node 20:
50
/ \
30 70
\ / \
40 60 80
the code fit here, because we're in the right subtree.
so once update 30 with 40 (root = root -> right) the situation would be this:
50
x \
40 70
/ \
60 80
then the piece of code #Bonje Fir wrote us, would fit perfectly:
tmp->Father->Left = root
because for sure, we're assigning 40 to the left son of the father of the original root. (because we're in the left subtree.)
50
/ \
40 70
/ \
60 80
So i made a little change to correct this logic problem, and make it work both in the right and left subtree.
else if (!root->Left) {
TreeNode *tmp = root;
root = tmp->Right;
root->Father = tmp->Father; //linking the father to the new son
//we need also to connect the son with the father, but first
//we need to know in which subtree we're in.
if (root->Father->Right == tmp) //if we're in the right subtree
tmp->Father->Right = root;
else ////if we're in the left subtree
tmp->Father->Left = root;
delete tmp;
std::cout << "Erased!" << std::endl;
}
i took advantage of the fact i didnt erase my root, once assigned the new one, so the father of the root still points to the old root.
(same speech for the opposite case.)

AVL Tree implementation c++

So I've posted about this recently, but I'm still at a loss for what is going wrong. Specifically, I can't seem to figure out what's causing my AVL Tree to take so long to sort. I read in a file of 500,000 random, unsorted numbers to sort by using a vector in a for loop to feed the tree the numbers one at a time. Now, I've also tested using a normal BST, as someone mentioned that having to create so many nodes one at a time might be why it's taking so long, but that completed in only 5 seconds, with only 12,164 nodes skipped due to being duplicates. My AVL Tree is taking upwards of 3 hours just to sort half the list, so something must be going wrong. Can anyone figure out what it is? As far as I know, the rebalancing and insertion logic is correct, because whenever I ran a bunch of test cases on it they all came out fine. I can't seem to track down where the problem is. Here's my full code for anyone that wants to check it out. Main is kind of a mess right now because of all the stuff I've included for testing purposes (like the tracking loop), but most of that will be gone in the final version.
EDIT:
This question has been answered.
#include <iostream>
#include<iomanip>
#include <time.h>
#include <vector>
#include <fstream>
using namespace std;
vector<int> numbers;
struct node
{
public:
int data, height;
node *leftChild, *rightChild;
};
node* root = NULL;
int findMin(node *p) // finds the smallest node in the tree
{
while (p->leftChild != NULL)
p = p->leftChild;
return p->data;
}
int findMax(node *p) // finds the largest node in the tree
{
while(p->rightChild != NULL)
p = p->rightChild;
return p->data;
}
int max(int a, int b) // gets the max of two integers
{
if(a > b)
return a;
else
return b;
}
int height(node *p) // gets the height of the tree
{
if(p == NULL)
return -1;
else
{
p->height = max(height(p->leftChild), height(p->rightChild)) + 1;
}
return p->height;
}
node* newNode(int element) // helper function to return a new node with empty subtrees
{
node* newPtr = new node;
newPtr->data = element;
newPtr->leftChild = NULL;
newPtr->rightChild = NULL;
newPtr->height = 1;
return newPtr;
}
node* rightRotate(node* p) // function to right rotate a tree rooted at p
{
node* child = p->leftChild; // rotate the tree
p->leftChild = child->rightChild;
child->rightChild = p;
// update the height for the nodes
p->height = height(p);
child->height = height(child);
// return new root
return child;
}
node* leftRotate(node* p) // function to left rotate a tree rooted at p
{
node* child = p->rightChild; // perform the rotation
p->rightChild = child->leftChild;
child->leftChild = p;
// update the heights for the nodes
p->height = height(p);
child->height = height(child);
// return new root
return child;
}
int getBalance(node *p)
{
if(p == NULL)
return 0;
else
return height(p->leftChild) - height(p->rightChild);
}
// recursive version of BST insert to insert the element in a sub tree rooted with root
// which returns new root of subtree
node* insert(node*& p, int element)
{
// perform the normal BST insertion
if(p == NULL) // if the tree is empty
return(newNode(element));
if(element < p->data)
{
p->leftChild = insert(p->leftChild, element);
}
else
{
p->rightChild = insert(p->rightChild, element);
}
// update the height for this node
p->height = height(p);
// get the balance factor to see if the tree is unbalanced
int balance = getBalance(p);
// the tree is unbalanced, there are 4 different types of rotation to make
// Single Right Rotation (Left Left Case)
if(balance > 1 && element < p->leftChild->data)
{
return rightRotate(p);
}
// Single Left Rotation (Right Right Case)
if(balance < -1 && element > p->rightChild->data)
{
return leftRotate(p);
}
// Left Right Rotation (double left rotation)
if(balance > 1 && element > p->leftChild->data)
{
p->leftChild = leftRotate(p->leftChild);
return rightRotate(p);
}
// Right Left Rotation
if(balance < -1 && element < p->rightChild->data)
{
p->rightChild = rightRotate(p->rightChild);
return leftRotate(p);
}
// cout << "Height: " << n->height << endl;
// return the unmodified root pointer in the case that the tree does not become unbalanced
return p;
}
void inorder(node *p)
{
if(p != NULL)
{
inorder(p->leftChild);
cout << p->data << ", ";
inorder(p->rightChild);
}
}
void preorder(node *p)
{
if(p != NULL)
{
cout << p->data << ", ";
preorder(p->leftChild);
preorder(p->rightChild);
}
}
void print(node* root)
{
/*cout << "Min Value: " << findMin(root) << endl;
cout << "Max Value: " << findMax(root) << endl;
cout << "Pre Order: ";
preorder(root); */
cout << endl << "Inorder: ";
inorder(root);
cout << endl << endl << endl << endl;
}
void read()
{
int num;
ifstream file_save("data.txt");
if(file_save.is_open())
{
while(!file_save.eof())
{
file_save >> num;
numbers.push_back(num);
}
file_save.close();
}
else
{
cout << "Error in opening file!!" << endl;
}
}
int main()
{
double duration;
time_t begin = time(0);
read();
int x = 0;
int track = 0;
for (std::vector<int>::const_iterator i = numbers.begin(); i != numbers.begin() + 100000; ++i)
{
root = insert(root, numbers[x]);
x++;
track++;
if( (track % 10000) == 0)
{
cout << track << " iterations" << endl;
time_t now = time(0);
cout << now - begin << " seconds" << endl;
}
}
time_t end = time(0);
duration = end - begin;
// print(root);
cout << "The algorithm took " << duration << " seconds to complete." << endl;
return 0;
}
There are many problems with this code.
while(eof) is wrong.
The main loop expects exactly 100000 elements.
All key comparisons are exact (<, >). There are no rotations performed when a duplicate element is inserted. Thus a tree of identical elements will not be balanced at all.
The height of an empty tree is hardcoded to -1, but the height of a single-node three is initially set to 1, thus violating the invariant height(node) = 1+max(height(node->leftChild))+height(node->rightChild)).
height traverses the entire tree every time it is called, thus making insertion O(n).
So, it seems to me that the reason that it was taking so long was because of too many recursive calls all over the place. This modified code has less recursive calls and thus bogs down the CPU with less stacks to have to process. At least, that's what I'm getting out of this.
void newHeight(node* p)
{
double leftHeight = height(p->leftChild);
double rightHeight = height(p->rightChild);
if(leftHeight > rightHeight)
p->height = leftHeight;
else
p->height = rightHeight;
}
node* rotateright(node* p) // the right rotation round p
{
node* q = p->leftChild;
p->leftChild = q->rightChild;
q->rightChild = p;
newHeight(p);
newHeight(q);
return q;
}
node* rotateleft(node* q) // the left rotation round q
{
node* p = q->rightChild;
q->rightChild = p->leftChild;
p->leftChild = q;
newHeight(q);
newHeight(p);
return p;
}
node* rebalance(node* p) // p node balance
{
newHeight(p);
if( getBalance(p)==2 )
{
if( getBalance(p->rightChild) < 0 )
p->rightChild = rotateright(p->rightChild);
return rotateleft(p);
}
if (getBalance(p)==-2 )
{
if( getBalance(p->leftChild) > 0 )
p->leftChild = rotateleft(p->leftChild);
return rotateright(p);
}
return p; // no balance needed
}
node* insert(node* p, int element) // k key insertion in the tree with p root
{
if(!p) return newNode(element);
if(element < p->data)
p->leftChild = insert(p->leftChild, element);
else
p->rightChild = insert(p->rightChild, element);
return rebalance(p);
}

Generating a balanced binary tree

I am not sure if this is doing what it is supposed to do.
My goal is to generate a balanced binary tree, from a set of values.
Please let me know if this is correct.
NOTE: NOT a balanced binary search tree, just balanced binary tree.
int heightPrivate(nodePtr node)
{
if (node == NULL)
return -1;
return 1 + std::max(heightPrivate(node->left), heightPrivate(node->right));
}
void addNodePrivate(nodePtr node, int val)
{
if (root == NULL)
{
root = new BTNode;
root->data = val;
root->left = root->right = NULL;
}
else
{
if (node->left == NULL)
{
node->left = new BTNode;
node->left->data = val;
node->left->left = node->left->right = NULL;
}
else if (node->right == NULL)
{
node->right = new BTNode;
node->right->data = val;
node->right->left = node->right->right = NULL;
}
else
{
int lheight = heightPrivate(node->left);
int rheight = heightPrivate(node->right);
if (lheight < rheight)
addNodePrivate(node->left, val);
else if (rheight < lheight)
addNodePrivate(node->right, val);
else
addNodePrivate(node->left, val);
}
}
}
void printPostorderPrivate(nodePtr p, int indent=0)
{
if(p != NULL) {
if(p->left) printPostorderPrivate(p->left, indent+4);
if(p->right) printPostorderPrivate(p->right, indent+4);
if (indent) {
std::cout << std::setw(indent) << ' ';
}
std::cout<< p->data << " \n ";
}
}
In main...
int main()
{
BTree tree;
tree.addNode(1);
tree.addNode(2);
tree.addNode(3);
tree.addNode(4);
tree.addNode(5);
tree.addNode(6);
tree.addNode(7);
tree.printPostorder();
The result I get is this:
7
4
6
2
5
3
1
The children of 2 are 4 and 5. The question is why is it 7 going on the next level.
The reason that 7 appears where it does is because in the addNodePrivate method checks the heights of the two child branches, and if they are equal it goes left.
So as you insert 7, when the program is at the root (node 1) it sees that the height of the left branch and height of the right branch are both equal to 1 (node 2 had children 5 and 4 but no grandchildren, and node 3 has child 6 and also no grandchildren), and so it goes left - down the branch with node 2.
To achieve what you want, you need to choose the branch which has the shortest path, so comparing the height of two branches is not enough.
Hope that helps, best of luck.

Printing Certain Elements In a Tree

I am required to make a program that will accept numbers from the user, the numbers are interpreted as a rating, for a game. The point of this program is to accept the numbers and make uneven teams in quality not quantity.
For example:
The user enters the numbers: 25, 50, 63, 80.
The bad team would be: 25, 50.
The good team would be: 63, 90.
I am using a binary search tree so the time is O(nlogn).
My code so far accepts numbers, puts them into a tree, determines if the teams are equal in numbers, but does not print them properly. I suspect the problem is either in my int main() when I call the print function or the actual print function.
#include <iostream>
using namespace std;
struct node
{
int data;
node *left;
node *right;
};
node *createNode(int data) // creates nodes
{
node *newNode = NULL;
newNode = new node;
newNode->data=data;
newNode->left=NULL;
newNode->right=NULL;
return newNode;
}
node *insert(int data, node **tree) // inserts them into a tree
{
node *newNode=NULL;
if (*tree==NULL)
{
newNode = createNode(data);
*tree = newNode;
}
else if (data<(*tree)->data)
{
if ((*tree)->left==NULL)
{
newNode = createNode(data);
(*tree)->left=newNode;
}
else
{
newNode = insert(data, &((*tree)->left));
}
}
else
{
if ((*tree)->right==NULL)
{
newNode = createNode(data);
(*tree)->right = newNode;
}
else
newNode = insert(data, &((*tree)->right));
}
return newNode;
}
void destroy(node *tree) // destroy tree at the end
{
if (tree!=NULL)
{
if (tree->left!=NULL)
destroy(tree->left);
if (tree->right!=NULL)
destroy(tree->right);
delete tree;
}
}
int treeHeight(node *tree, int counter) // finds the height of the tree
{
if (tree==NULL) // if there is nothing in the tree, return 0
return counter;
int leftcount, rightcount; // leftcount counts the left side of the tree and right count counts the right side of the tree
counter++; // counter should go up by 1 for every node
rightcount = treeHeight(tree->right, counter); // searches the right side
leftcount = treeHeight(tree->left, counter); // searches the left side
if (rightcount > leftcount) // return the height
return rightcount;
else
return leftcount;
}
void printNode(node *Node)
{
if (Node!=NULL)
{
cout << Node->data << ", ";
}
}
void printLeft(node *tree)
{
if (tree!=NULL)
{
printLeft(tree->left);
printNode(tree);
}
}
void printRight(node *tree)
{
if (tree!=NULL)
{
printRight(tree->right);
printNode(tree);
}
}
int main()
{
node *root = NULL;
node *current = NULL;
int value;
while (true)
{
cout << "Enter a rating where the bigger the number, the better (zero to quit): ";
cin >> value;
if (value==0)
break;
current = insert(value, &root);
}
int height = treeHeight(root, 0); // call the function and display the height of the tree
if (height==0)
{
cout << "\nYou did not enter any players, the game cannot happen now. Thanks a lot." << endl;
return 0;
}
else if (height%2!=0)
{
cout << "Uneven amount of player, add one more player to make the teams equal in quantity" << endl;
cout << "Enter a rating: ";
cin >> value;
current = insert(value, &root);
}
cout << "\nThe bad team is: " ;
printLeft(root); // ???
cout << "\nThe good team is: ";
printRight(root); // ???
destroy(root);
return 0;
}
Firstly, you must have misunderstood something. Binary Search Tree with this input: 25, 50, 63, 80 would not be divided in a way you wrote. Look. it would be constructed like that:
25
\
50
\
63
\
80
Not like that:
25
/ \
50 63
\
80
And not like that
50
/ \
25 63
\
80
I would advise you to go and simulate insertions to BST: BST simulator online
Now, about your code. Functions printLeft and printRight print only elements from left-most and right-most branch of the tree (and printLeft do it top-bottom while printRight do it bottom-top). That's why they give results 25 and 80, 63, 50, 25. I don't know exactly what your BST should be divided into "bad" and "good" team, but let's suppose that "bad" team consists of root node and left subtree thile "good" team consists of only right subtree (wihtour root node). Then you could use this function:
void printSubTree(node *tree) {
if (tree) {
printSubTree(tree->left);
printNode(tree);
printSubTree(tree->right);
}
}
And modify your code to:
if (root) {
cout << "\nThe bad team is: " ;
printNode(root);
printSubTree(root->left);
cout << "\nThe good team is: ";
printSubTree(root->right);
}

Print a binary tree in a pretty way

Just wondering if I can get some tips on printing a pretty binary tree in the form of:
5
10
11
7
6
3
4
2
Right now what it prints is:
2
4
3
6
7
11
10
5
I know that my example is upside down from what I'm currently printing, which it doesn't matter if I print from the root down as it currently prints. Any tips are very appreciated towards my full question:
How do I modify my prints to make the tree look like a tree?
//Binary Search Tree Program
#include <iostream>
#include <cstdlib>
#include <queue>
using namespace std;
int i = 0;
class BinarySearchTree
{
private:
struct tree_node
{
tree_node* left;
tree_node* right;
int data;
};
tree_node* root;
public:
BinarySearchTree()
{
root = NULL;
}
bool isEmpty() const { return root==NULL; }
void print_inorder();
void inorder(tree_node*);
void print_preorder();
void preorder(tree_node*);
void print_postorder();
void postorder(tree_node*);
void insert(int);
void remove(int);
};
// Smaller elements go left
// larger elements go right
void BinarySearchTree::insert(int d)
{
tree_node* t = new tree_node;
tree_node* parent;
t->data = d;
t->left = NULL;
t->right = NULL;
parent = NULL;
// is this a new tree?
if(isEmpty()) root = t;
else
{
//Note: ALL insertions are as leaf nodes
tree_node* curr;
curr = root;
// Find the Node's parent
while(curr)
{
parent = curr;
if(t->data > curr->data) curr = curr->right;
else curr = curr->left;
}
if(t->data < parent->data)
{
parent->left = t;
}
else
{
parent->right = t;
}
}
}
void BinarySearchTree::remove(int d)
{
//Locate the element
bool found = false;
if(isEmpty())
{
cout<<" This Tree is empty! "<<endl;
return;
}
tree_node* curr;
tree_node* parent;
curr = root;
while(curr != NULL)
{
if(curr->data == d)
{
found = true;
break;
}
else
{
parent = curr;
if(d>curr->data) curr = curr->right;
else curr = curr->left;
}
}
if(!found)
{
cout<<" Data not found! "<<endl;
return;
}
// 3 cases :
// 1. We're removing a leaf node
// 2. We're removing a node with a single child
// 3. we're removing a node with 2 children
// Node with single child
if((curr->left == NULL && curr->right != NULL) || (curr->left != NULL && curr->right == NULL))
{
if(curr->left == NULL && curr->right != NULL)
{
if(parent->left == curr)
{
parent->left = curr->right;
delete curr;
}
else
{
parent->right = curr->left;
delete curr;
}
}
return;
}
//We're looking at a leaf node
if( curr->left == NULL && curr->right == NULL)
{
if(parent->left == curr)
{
parent->left = NULL;
}
else
{
parent->right = NULL;
}
delete curr;
return;
}
//Node with 2 children
// replace node with smallest value in right subtree
if (curr->left != NULL && curr->right != NULL)
{
tree_node* chkr;
chkr = curr->right;
if((chkr->left == NULL) && (chkr->right == NULL))
{
curr = chkr;
delete chkr;
curr->right = NULL;
}
else // right child has children
{
//if the node's right child has a left child
// Move all the way down left to locate smallest element
if((curr->right)->left != NULL)
{
tree_node* lcurr;
tree_node* lcurrp;
lcurrp = curr->right;
lcurr = (curr->right)->left;
while(lcurr->left != NULL)
{
lcurrp = lcurr;
lcurr = lcurr->left;
}
curr->data = lcurr->data;
delete lcurr;
lcurrp->left = NULL;
}
else
{
tree_node* tmp;
tmp = curr->right;
curr->data = tmp->data;
curr->right = tmp->right;
delete tmp;
}
}
return;
}
}
void BinarySearchTree::print_postorder()
{
postorder(root);
}
void BinarySearchTree::postorder(tree_node* p)
{
if(p != NULL)
{
if(p->left) postorder(p->left);
if(p->right) postorder(p->right);
cout<<" "<<p->data<<"\n ";
}
else return;
}
int main()
{
BinarySearchTree b;
int ch,tmp,tmp1;
while(1)
{
cout<<endl<<endl;
cout<<" Binary Search Tree Operations "<<endl;
cout<<" ----------------------------- "<<endl;
cout<<" 1. Insertion/Creation "<<endl;
cout<<" 2. Printing "<<endl;
cout<<" 3. Removal "<<endl;
cout<<" 4. Exit "<<endl;
cout<<" Enter your choice : ";
cin>>ch;
switch(ch)
{
case 1 : cout<<" Enter Number to be inserted : ";
cin>>tmp;
b.insert(tmp);
i++;
break;
case 2 : cout<<endl;
cout<<" Printing "<<endl;
cout<<" --------------------"<<endl;
b.print_postorder();
break;
case 3 : cout<<" Enter data to be deleted : ";
cin>>tmp1;
b.remove(tmp1);
break;
case 4:
return 0;
}
}
}
In order to pretty-print a tree recursively, you need to pass two arguments to your printing function:
The tree node to be printed, and
The indentation level
For example, you can do this:
void BinarySearchTree::postorder(tree_node* p, int indent=0)
{
if(p != NULL) {
if(p->left) postorder(p->left, indent+4);
if(p->right) postorder(p->right, indent+4);
if (indent) {
std::cout << std::setw(indent) << ' ';
}
cout<< p->data << "\n ";
}
}
The initial call should be postorder(root);
If you would like to print the tree with the root at the top, move cout to the top of the if.
void btree::postorder(node* p, int indent)
{
if(p != NULL) {
if(p->right) {
postorder(p->right, indent+4);
}
if (indent) {
std::cout << std::setw(indent) << ' ';
}
if (p->right) std::cout<<" /\n" << std::setw(indent) << ' ';
std::cout<< p->key_value << "\n ";
if(p->left) {
std::cout << std::setw(indent) << ' ' <<" \\\n";
postorder(p->left, indent+4);
}
}
}
With this tree:
btree *mytree = new btree();
mytree->insert(2);
mytree->insert(1);
mytree->insert(3);
mytree->insert(7);
mytree->insert(10);
mytree->insert(2);
mytree->insert(5);
mytree->insert(8);
mytree->insert(6);
mytree->insert(4);
mytree->postorder(mytree->root);
Would lead to this result:
It's never going to be pretty enough, unless one does some backtracking to re-calibrate the display output. But one can emit pretty enough binary trees efficiently using heuristics: Given the height of a tree, one can guess what the expected width and setw of nodes at different depths.
There are a few pieces needed to do this, so let's start with the higher level functions first to provide context.
The pretty print function:
// create a pretty vertical tree
void postorder(Node *p)
{
int height = getHeight(p) * 2;
for (int i = 0 ; i < height; i ++) {
printRow(p, height, i);
}
}
The above code is easy. The main logic is in the printRow function. Let's delve into that.
void printRow(const Node *p, const int height, int depth)
{
vector<int> vec;
getLine(p, depth, vec);
cout << setw((height - depth)*2); // scale setw with depth
bool toggle = true; // start with left
if (vec.size() > 1) {
for (int v : vec) {
if (v != placeholder) {
if (toggle)
cout << "/" << " ";
else
cout << "\\" << " ";
}
toggle = !toggle;
}
cout << endl;
cout << setw((height - depth)*2);
}
for (int v : vec) {
if (v != placeholder)
cout << v << " ";
}
cout << endl;
}
getLine() does what you'd expect: it stores all nodes with a given equal depth into vec. Here's the code for that:
void getLine(const Node *root, int depth, vector<int>& vals)
{
if (depth <= 0 && root != nullptr) {
vals.push_back(root->val);
return;
}
if (root->left != nullptr)
getLine(root->left, depth-1, vals);
else if (depth-1 <= 0)
vals.push_back(placeholder);
if (root->right != nullptr)
getLine(root->right, depth-1, vals);
else if (depth-1 <= 0)
vals.push_back(placeholder);
}
Now back to printRow(). For each line, we set the stream width based on how deep we are in the binary tree. This formatting will be nice because, typically, the deeper you go, the more width is needed. I say typically because in degenerate trees, this wouldn't look as pretty. As long as the tree is roughly balanced and smallish (< 20 items), it should turn out fine.
A placeholder is needed to align the '/' and '\' characters properly. So when a row is obtained via getLine(), we insert the placeholder if there isn't any node present at the specified depth. The placeholder can be set to anything like (1<<31) for example. Obviously, this isn't robust because the placeholder could be a valid node value. If a coder's got spunk and is only dealing with decimals, one could modify the code to emit decimal-converted strings via getLine() and use a placeholder like "_". (Unfortunately, I'm not such a coder :P)
The result for the following items inserted in order: 8, 12, 4, 2, 5, 15 is
8
/ \
4 12
/ \ \
2 5 15
getHeight() is left to the reader as an exercise. :)
One could even get prettier results by retroactively updating the setw of shallow nodes based on the number of items in deeper nodes.
That too is left to the reader as an exercise.
#include <stdio.h>
#include <stdlib.h>
struct Node
{
struct Node *left,*right;
int val;
} *root=NULL;
int rec[1000006];
void addNode(int,struct Node*);
void printTree(struct Node* curr,int depth)
{
int i;
if(curr==NULL)return;
printf("\t");
for(i=0;i<depth;i++)
if(i==depth-1)
printf("%s\u2014\u2014\u2014",rec[depth-1]?"\u0371":"\u221F");
else
printf("%s ",rec[i]?"\u23B8":" ");
printf("%d\n",curr->val);
rec[depth]=1;
printTree(curr->left,depth+1);
rec[depth]=0;
printTree(curr->right,depth+1);
}
int main()
{
root=(struct Node*)malloc(sizeof(struct Node));
root->val=50;
//addNode(50,root);
addNode(75,root); addNode(25,root);
addNode(15,root); addNode(30,root);
addNode(100,root); addNode(60,root);
addNode(27,root); addNode(31,root);
addNode(101,root); addNode(99,root);
addNode(5,root); addNode(61,root);
addNode(55,root); addNode(20,root);
addNode(0,root); addNode(21,root);
//deleteNode(5,root);
printTree(root,0);
return 0;
}
void addNode(int v,struct Node* traveller)
{
struct Node *newEle=(struct Node*)malloc(sizeof(struct Node));
newEle->val=v;
for(;;)
{
if(v<traveller->val)
{
if(traveller->left==NULL){traveller->left=newEle;return;}
traveller=traveller->left;
}
else if(v>traveller->val)
{
if(traveller->right==NULL){traveller->right=newEle;return;}
traveller=traveller->right;
}
else
{
printf("%d Input Value is already present in the Tree !!!\n",v);
return;
}
}
}
Hope, you find it pretty...
Output:
50
ͱ———25
⎸ ͱ———15
⎸ ⎸ ͱ———5
⎸ ⎸ ⎸ ͱ———0
⎸ ⎸ ∟———20
⎸ ⎸ ∟———21
⎸ ∟———30
⎸ ͱ———27
⎸ ∟———31
∟———75
ͱ———60
⎸ ͱ———55
⎸ ∟———61
∟———100
ͱ———99
∟———101
//Binary tree (pretty print):
// ________________________50______________________
// ____________30 ____________70__________
// ______20____ 60 ______90
// 10 15 80
// prettyPrint
public static void prettyPrint(BTNode node) {
// get height first
int height = heightRecursive(node);
// perform level order traversal
Queue<BTNode> queue = new LinkedList<BTNode>();
int level = 0;
final int SPACE = 6;
int nodePrintLocation = 0;
// special node for pushing when a node has no left or right child (assumption, say this node is a node with value Integer.MIN_VALUE)
BTNode special = new BTNode(Integer.MIN_VALUE);
queue.add(node);
queue.add(null); // end of level 0
while(! queue.isEmpty()) {
node = queue.remove();
if (node == null) {
if (!queue.isEmpty()) {
queue.add(null);
}
// start of new level
System.out.println();
level++;
} else {
nodePrintLocation = ((int) Math.pow(2, height - level)) * SPACE;
System.out.print(getPrintLine(node, nodePrintLocation));
if (level < height) {
// only go till last level
queue.add((node.left != null) ? node.left : special);
queue.add((node.right != null) ? node.right : special);
}
}
}
}
public void prettyPrint() {
System.out.println("\nBinary tree (pretty print):");
prettyPrint(root);
}
private static String getPrintLine(BTNode node, int spaces) {
StringBuilder sb = new StringBuilder();
if (node.data == Integer.MIN_VALUE) {
// for child nodes, print spaces
for (int i = 0; i < 2 * spaces; i++) {
sb.append(" ");
}
return sb.toString();
}
int i = 0;
int to = spaces/2;
for (; i < to; i++) {
sb.append(' ');
}
to += spaces/2;
char ch = ' ';
if (node.left != null) {
ch = '_';
}
for (; i < to; i++) {
sb.append(ch);
}
String value = Integer.toString(node.data);
sb.append(value);
to += spaces/2;
ch = ' ';
if (node.right != null) {
ch = '_';
}
for (i += value.length(); i < to; i++) {
sb.append(ch);
}
to += spaces/2;
for (; i < to; i++) {
sb.append(' ');
}
return sb.toString();
}
private static int heightRecursive(BTNode node) {
if (node == null) {
// empty tree
return -1;
}
if (node.left == null && node.right == null) {
// leaf node
return 0;
}
return 1 + Math.max(heightRecursive(node.left), heightRecursive(node.right));
}
If your only need is to visualize your tree, a better method would be to output it into a dot format and draw it with grapviz.
You can look at dot guide for more information abt syntax etc
Here's yet another C++98 implementation, with tree like output.
Sample output:
PHP
└── is
├── minor
│ └── perpetrated
│ └── whereas
│ └── skilled
│ └── perverted
│ └── professionals.
└── a
├── evil
│ ├── incompetent
│ │ ├── insidious
│ │ └── great
│ └── and
│ ├── created
│ │ └── by
│ │ └── but
│ └── amateurs
└── Perl
The code:
void printTree(Node* root)
{
if (root == NULL)
{
return;
}
cout << root->val << endl;
printSubtree(root, "");
cout << endl;
}
void printSubtree(Node* root, const string& prefix)
{
if (root == NULL)
{
return;
}
bool hasLeft = (root->left != NULL);
bool hasRight = (root->right != NULL);
if (!hasLeft && !hasRight)
{
return;
}
cout << prefix;
cout << ((hasLeft && hasRight) ? "├── " : "");
cout << ((!hasLeft && hasRight) ? "└── " : "");
if (hasRight)
{
bool printStrand = (hasLeft && hasRight && (root->right->right != NULL || root->right->left != NULL));
string newPrefix = prefix + (printStrand ? "│ " : " ");
cout << root->right->val << endl;
printSubtree(root->right, newPrefix);
}
if (hasLeft)
{
cout << (hasRight ? prefix : "") << "└── " << root->left->val << endl;
printSubtree(root->left, prefix + " ");
}
}
Here's a little example for printing out an array based heap in tree form. It would need a little adjusting to the algorithm for bigger numbers. I just made a grid on paper and figured out what space index each node would be to look nice, then noticed there was a pattern to how many spaces each node needed based on its parent's number of spaces and the level of recursion as well as how tall the tree is. This solution goes a bit beyond just printing in level order and satisfies the "beauty" requirement.
#include <iostream>
#include <vector>
static const int g_TerminationNodeValue = -999;
class HeapJ
{
public:
HeapJ(int* pHeapArray, int numElements)
{
m_pHeapPointer = pHeapArray;
m_numElements = numElements;
m_treeHeight = GetTreeHeight(1);
}
void Print()
{
m_printVec.clear();
int initialIndex = 0;
for(int i=1; i<m_treeHeight; ++i)
{
int powerOfTwo = 1;
for(int j=0; j<i; ++j)
{
powerOfTwo *= 2;
}
initialIndex += powerOfTwo - (i-1);
}
DoPrintHeap(1,0,initialIndex);
for(size_t i=0; i<m_printVec.size(); ++i)
{
std::cout << m_printVec[i] << '\n' << '\n';
}
}
private:
int* m_pHeapPointer;
int m_numElements;
int m_treeHeight;
std::vector<std::string> m_printVec;
int GetTreeHeight(int index)
{
const int value = m_pHeapPointer[index-1];
if(value == g_TerminationNodeValue)
{
return -1;
}
const int childIndexLeft = 2*index;
const int childIndexRight = childIndexLeft+1;
int valLeft = 0;
int valRight = 0;
if(childIndexLeft <= m_numElements)
{
valLeft = GetTreeHeight(childIndexLeft);
}
if(childIndexRight <= m_numElements)
{
valRight = GetTreeHeight(childIndexRight);
}
return std::max(valLeft,valRight)+1;
}
void DoPrintHeap(int index, size_t recursionLevel, int numIndents)
{
const int value = m_pHeapPointer[index-1];
if(value == g_TerminationNodeValue)
{
return;
}
if(m_printVec.size() == recursionLevel)
{
m_printVec.push_back(std::string(""));
}
const int numLoops = numIndents - (int)m_printVec[recursionLevel].size();
for(int i=0; i<numLoops; ++i)
{
m_printVec[recursionLevel].append(" ");
}
m_printVec[recursionLevel].append(std::to_string(value));
const int childIndexLeft = 2*index;
const int childIndexRight = childIndexLeft+1;
const int exponent = m_treeHeight-(recursionLevel+1);
int twoToPower = 1;
for(int i=0; i<exponent; ++i)
{
twoToPower *= 2;
}
const int recursionAdjust = twoToPower-(exponent-1);
if(childIndexLeft <= m_numElements)
{
DoPrintHeap(childIndexLeft, recursionLevel+1, numIndents-recursionAdjust);
}
if(childIndexRight <= m_numElements)
{
DoPrintHeap(childIndexRight, recursionLevel+1, numIndents+recursionAdjust);
}
}
};
const int g_heapArraySample_Size = 14;
int g_heapArraySample[g_heapArraySample_Size] = {16,14,10,8,7,9,3,2,4,1,g_TerminationNodeValue,g_TerminationNodeValue,g_TerminationNodeValue,0};
int main()
{
HeapJ myHeap(g_heapArraySample,g_heapArraySample_Size);
myHeap.Print();
return 0;
}
/* output looks like this:
16
14 10
8 7 9 3
2 4 1 0
*/
Foreword
Late late answer and its in Java, but I'd like to add mine to the record because I found out how to do this relatively easily and the way I did it is more important. The trick is to recognize that what you really want is for none of your sub-trees to be printed directly under your root/subroot nodes (in the same column). Why you might ask? Because it Guarentees that there are no spacing problems, no overlap, no possibility of the left subtree and right subtree ever colliding, even with superlong numbers. It auto adjusts to the size of your node data. The basic idea is to have the left subtree be printed totally to the left of your root and your right subtree is printed totally to the right of your root.
An anaology of how I though about this problem
A good way to think about it is with Umbrellas, Imagine first that you are outside with a large umbrella, you represent the root and your Umbrella and everything under it is the whole tree. think of your left subtree as a short man (shorter than you anyway) with a smaller umbrella who is on your left under your large umbrella. Your right subtree is represented by a similar man with a similarly smaller umbrella on your right side. Imagine that if the umbrellas of the short men ever touch, they get angry and hit each other (bad overlap). You are the root and the men beside you are your subtrees. You must be exactly in the middle of their umbrellas (subtrees) to break up the two men and ensure they never bump umbrellas. The trick is to then imagine this recursively, where each of the two men each have their own two smaller people under their umbrella (children nodes) with ever smaller umbrellas (sub-subtrees and so-on) that they need to keep apart under their umbrella (subtree), They act as sub-roots. Fundamentally, thats what needs to happen to 'solve' the general problem when printing binary trees, subtree overlap. To do this, you simply need to think about how you would 'print' or 'represent' the men in my anaolgy.
My implementation, its limitations and its potential
Firstly the only reason my code implementation takes in more parameters than should be needed (currentNode to be printed and node level) is because I can't easily move a line up in console when printing, so I have to map my lines first and print them in reverse. To do this I made a lineLevelMap that mapped each line of the tree to it's output (this might be useful for the future as a way to easily gather every line of the tree and also print it out at the same time).
//finds the height of the tree beforehand recursively, left to reader as exercise
int height = TreeHeight(root);
//the map that uses the height of the tree to detemrine how many entries it needs
//each entry maps a line number to the String of the actual line
HashMap<Integer,String> lineLevelMap = new HashMap<>();
//initialize lineLevelMap to have the proper number of lines for our tree
//printout by starting each line as the empty string
for (int i = 0; i < height + 1; i++) {
lineLevelMap.put(i,"");
}
If I could get ANSI escape codes working in the java console (windows ugh) I could simply print one line upwards and I would cut my parameter count by two because I wouldn't need to map lines or know the depth of the tree beforehand. Regardless here is my code that recurses in an in-order traversal of the tree:
public int InOrderPrint(CalcTreeNode currentNode, HashMap<Integer,String>
lineLevelMap, int level, int currentIndent){
//traverse left case
if(currentNode.getLeftChild() != null){
//go down one line
level--;
currentIndent =
InOrderPrint(currentNode.getLeftChild(),lineLevelMap,level,currentIndent);
//go up one line
level++;
}
//find the string length that already exists for this line
int previousIndent = lineLevelMap.get(level).length();
//create currentIndent - previousIndent spaces here
char[] indent = new char[currentIndent-previousIndent];
Arrays.fill(indent,' ');
//actually append the nodeData and the proper indent to add on to the line
//correctly
lineLevelMap.put(level,lineLevelMap.get(level).concat(new String(indent) +
currentNode.getData()));
//update the currentIndent for all lines
currentIndent += currentNode.getData().length();
//traverse right case
if (currentNode.getRightChild() != null){
//go down one line
level--;
currentIndent =
InOrderPrint(currentNode.getRightChild(),lineLevelMap,level,currentIndent);
//go up one line
level++;
}
return currentIndent;
}
To actually print this Tree to console in java, just use the LineMap that we generated. This way we can print the lines right side up
for (int i = height; i > -1; i--) {
System.out.println(lineLevelMap.get(i));
}
How it all really works
The InorderPrint sub function does all the 'work' and can recursively print out any Node and it's subtrees properly. Even better, it spaces them evenly and you can easily modify it to space out all nodes equally (just make the Nodedata equal or make the algorithim think it is). The reason it works so well is because it uses the Node's data length to determine where the next indent should be. This assures that the left subtree is always printed BEFORE the root and the right subtree, thus if you ensure this recursively, no left node is printed under it's root nor its roots root and so-on with the same thing true for any right node. Instead the root and all subroots are directly in the middle of their subtrees and no space is wasted.
An example output with an input of 3 + 2 looks like in console is:
And an example of 3 + 4 * 5 + 6 is:
And finally an example of ( 3 + 4 ) * ( 5 + 6 ) note the parenthesis is:
Ok but why Inorder?
The reason an Inorder traversal works so well is because it Always prints the leftmost stuff first, then the root, then the rightmost stuff. Exactly how we want our subtrees to be: everything to the left of the root is printed to the left of the root, everything to the right is printed to the right. Inorder traversal naturally allows for this relationship, and since we print lines and make indents based on nodeData, we don't need to worry about the length of our data. The node could be 20 characters long and it wouldn't affect the algorithm (although you might start to run out of actual screen space). The algorithm doesn't create any spacing between nodes but that can be easily implemented, the important thing is that they don't overlap.
Just to prove it for you (don't take my word for this stuff) here is an example with some quite long characters
As you can see, it simply adjusts based on the size of the data, No overlap! As long as your screen is big enough. If anyone ever figures out an easy way to print one line up in the java console (I'm all ears) This will become much much simpler, easy enough for almost anyone with basic knowledge of trees to understand and use, and the best part is there is no risk of bad overlapping errors.
Do an in-order traversal, descending to children before moving to siblings. At each level, that is when you descent to a child, increase the indent. After each node you output, print a newline.
Some psuedocode. Call Print with the root of your tree.
void PrintNode(int indent, Node* node)
{
while (--indent >= 0)
std::cout << " ";
std::cout << node->value() << "\n";
}
void PrintNodeChildren(int indent, Node* node)
{
for (int child = 0; child < node->ChildCount(); ++child)
{
Node* childNode = node->GetChild(child);
PrintNode(indent, childNode);
PrintNodeChildren(indent + 1, childNode);
}
}
void Print(Node* root)
{
int indent = 0;
PrintNode(indent, root);
PrintNodeChildren(indent + 1, root);
}
From your root, count the number of your left children. From the total number of left children, proceed with printing the root with the indention of the number of left children. Move to the next level of the tree with the decremented number of indention for the left child, followed by an initial two indentions for the right child. Decrement the indention of the left child based on its level and its parent with a double indention for its right sibling.
For an Array I find this much more concise. Merely pass in the array. Could be improved to handle very large numbers(long digit lengths). Copy and paste for c++ :)
#include <math.h>
using namespace std;
void printSpace(int count){
for (int x = 0; x<count; x++) {
cout<<"-";
}
}
void printHeap(int heap[], int size){
cout<<endl;
int height = ceil(log(size)+1); //+1 handle the last leaves
int width = pow(2, height)*height;
int index = 0;
for (int x = 0; x <= height; x++) { //for each level of the tree
for (int z = 0; z < pow(2, x); z++) { // for each node on that tree level
int digitWidth = 1;
if(heap[index] != 0) digitWidth = floor(log10(abs(heap[index]))) + 1;
printSpace(width/(pow(2,x))-digitWidth);
if(index<size)cout<<heap[index++];
else cout<<"-";
printSpace(width/(pow(2,x)));
}
cout<<endl;
}
}
Here is preorder routine that prints a general tree graph in a compact way:
void preOrder(Node* nd, bool newLine=false,int indent=0)
{
if(nd != NULL) {
if (newLine && indent) {
std::cout << "\n" << std::setw(indent) << ' '
} else if(newLine)
std::cout << "\n";
cout<< nd->_c;
vector<Node *> &edges=nd->getEdges();
int eSize=edges.size();
bool nwLine=false;
for(int i=0; i<eSize; i++) {
preOrder(edges[i],nwLine,indent+1);
nwLine=true;
}
}
}
int printGraph()
{
preOrder(root,true);
}
i have a easier code..........
consider a tree made of nodes of structure
struct treeNode{
treeNode *lc;
element data;
short int bf;
treeNode *rc;
};
Tree's depth can be found out using
int depth(treeNode *p){
if(p==NULL) return 0;
int l=depth(p->lc);
int r=depth(p->rc);
if(l>=r)
return l+1;
else
return r+1;
}
below gotoxy function moves your cursor to the desired position
void gotoxy(int x,int y)
{
printf("%c[%d;%df",0x1B,y,x);
}
Then Printing a Tree can be done as:
void displayTreeUpDown(treeNode * root,int x,int y,int px=0){
if(root==NULL) return;
gotoxy(x,y);
int a=abs(px-x)/2;
cout<<root->data.key;
displayTreeUpDown(root->lc,x-a,y+1,x);
displayTreeUpDown(root->rc,x+a,y+1,x);
}
which can be called using:
display(t,pow(2,depth(t)),1,1);
Here is my code. It prints very well,maybe its not perfectly symmetrical.
little description:
1st function - prints level by level (root lv -> leaves lv)
2nd function - distance from the beginning of new line
3rd function - prints nodes and calculates distance between two prints;
void Tree::TREEPRINT()
{
int i = 0;
while (i <= treeHeight(getroot())){
printlv(i);
i++;
cout << endl;
}
}
void Tree::printlv(int n){
Node* temp = getroot();
int val = pow(2, treeHeight(root) -n+2);
cout << setw(val) << "";
prinlv(temp, n, val);
}
void Tree::dispLV(Node*p, int lv, int d)
{
int disp = 2 * d;
if (lv == 0){
if (p == NULL){
cout << " x ";
cout << setw(disp -3) << "";
return;
}
else{
int result = ((p->key <= 1) ? 1 : log10(p->key) + 1);
cout << " " << p->key << " ";
cout << setw(disp - result-2) << "";
}
}
else
{
if (p == NULL&& lv >= 1){
dispLV(NULL, lv - 1, d);
dispLV(NULL, lv - 1, d);
}
else{
dispLV(p->left, lv - 1, d);
dispLV(p->right, lv - 1, d);
}
}
}
Input:
50-28-19-30-29-17-42-200-160-170-180-240-44-26-27
Output: https://i.stack.imgur.com/TtPXY.png
This code is written in C. It will basically print the tree "floor by floor".
Example of the output:
The function rb_tree_putchar_fd() can be replaced by a basic function that prints on screen, like std::cout << ... ;
SIZE_LEAF_DEBUG should be replaced by an int, and should be an even number. Use 6 for conveniance.
The function display() has one role: always print SIZE_LEAF_DEBUG characters on screen. I used '[' + 4 characters + ']' in my example. The four characters can be the string representation of an int for example.
//#include "rb_tree.h"
#define SIZE_LEAF_DEBUG 6
int rb_tree_depth(t_rb_node *root);
/*
** note: This debugging function will display the red/black tree in a tree
** fashion.
** RED nodes are displayed in red.
**
** note: The custom display func takes care of displaying the item of a node
** represented as a string of SIZE_LEAF_DEBUG characters maximum,
** padded with whitespaces if necessary. If item is null: the leaf is
** represented as "[null]"...
**
** note: the define SIZE_LEAF_DEBUG should be used by the display func.
** SIZE_LEAF_DEBUG should be an even number.
**
** note: Every node is represented by:
** - either whitespaces if NULL
** - or between squarred brackets a string representing the item.
*/
/*
** int max; //max depth of the rb_tree
** int current; //current depth while recursing
** int bottom; //current is trying to reach bottom while doing a bfs.
*/
typedef struct s_depth
{
int max;
int current;
int bottom;
} t_depth;
static void rb_tree_deb2(t_rb_node *node, t_depth depth, void (*display)())
{
int size_line;
int i;
i = 0;
size_line = (1 << (depth.max - ++depth.current)) * SIZE_LEAF_DEBUG;
if (!node)
{
while (i++ < size_line)
rb_tree_putchar_fd(' ', 1);
return ;
}
if (depth.current == depth.bottom)
{
while (i++ < (size_line - SIZE_LEAF_DEBUG) / 2)
rb_tree_putchar_fd(' ', 1);
if (node->color == RB_RED)
rb_tree_putstr_fd("\033[31m", 1);
display(node->item);
rb_tree_putstr_fd("\033[0m", 1);
while (i++ <= (size_line - SIZE_LEAF_DEBUG))
rb_tree_putchar_fd(' ', 1);
return ;
}
rb_tree_deb2(node->left, depth, display);
rb_tree_deb2(node->right, depth, display);
}
void rb_tree_debug(t_rb_node *root, void (*display)())
{
t_depth depths;
rb_tree_putstr_fd("\n===================================================="\
"===========================\n====================== BTREE DEBUG "\
"START ======================================\n", 1);
if (root && display)
{
depths.max = rb_tree_depth((t_rb_node*)root);
depths.current = 0;
depths.bottom = 0;
while (++depths.bottom <= depths.max)
{
rb_tree_deb2(root, depths, display);
rb_tree_putchar_fd('\n', 1);
}
}
else
rb_tree_putstr_fd("NULL ROOT, or NULL display func\n", 1);
rb_tree_putstr_fd("\n============================== DEBUG END ==========="\
"===========================\n==================================="\
"============================================\n\n\n", 1);
}