Insert into n-ary tree using Queue - c++

I have a simple n-ary (3 child nodes maximum) whereby the first node inserted will be the root. Before, I add any other node, I have to search the tree and insert as a child node from a previously inserted node, if a condition is meet.
My insertion methods is overloaded for first insertion and subsequent insertions.
I was able to insert the first node using this method:
void Tree::AddSkill(char* name, char* desc, int level)
{
Skill s(name, desc, level);
Node * newNode = new Node(s);
//newNode->aSkill = Skill(name, desc, level);
newNode->parent = NULL;
for (int i = 0; i<CHILD_MAX; i++)
{
newNode->children[i] = NULL;
}
if (this->root == NULL)
{
this->root = newNode;
}
else
{
this->root->parent = newNode;
newNode->children[0] = this->root;
this->root = newNode;
}
}
I'm having a few issues with subsequent insertion into the tree,
Here is the code I have so far:
void Tree::AddSkill(char* name, char* desc, int level, char* parentName)
{
if (this->root == NULL)
{
cout << "Error: no nodes in tree.\n";
return;
}
Node* node = NULL;
Skill s(name, desc, level);
Node * child = new Node(s);
while (root != NULL)
{
if (strcmp(child->aSkill.GetName(), parentName) == 0)
{
for (int i = 0; i < CHILD_MAX; i++)
{
if (node->children[i] == NULL)
{
child->aSkill = s;
child->parent = node;
node->children[i] = child;
return;
}
}
}
}
}
When I run the code through VS Debugger, the while loop in the second AddSkill method repeats endlessly.
I'm not so sure what I'm doing wrong or what concept I need to implement, any help will be appreciated.
P.S. This is an Homework (Not sure what the appropriate tag is).
Update:
I have tried to implement the overloaded AddSkill() using Queue.
This is what I've tried with it.
void SkillTree::AddSkill(char* name, char* desc, int level, char* parentName)
{
if (this->root == NULL)
{
cout << "Error: no nodes in tree.\n";
return;
}
queue<Node*> q;
q.push(this->root);
while (!q.empty())
{
Node * n = q.front();
q.pop();
if (strcmp(n->aSkill.GetName(), parentName) == 0)
{
for (int i = 0; i<CHILD_MAX; i++)
{
if (n->children[i] == NULL)
{
Skill s(name, desc, level);
Node * child = new Node(s);
//When I comment out the next 3 lines, program does not crash. Not sure what the problem is here.
child->aSkill = s;
child->parent = n;
n->children[i] = child;
return;
}
}
return;
}
for (int i = 0; i<CHILD_MAX; i++)
{
if (n->children[i] != NULL)
{
q.push(n->children[i]);
}
}
}
}
Skill Class
#include <iostream>
#include "Skill.h"
Skill::Skill()
{
name = NULL;
desc = NULL;
level = 0;
}
Skill::Skill(char* name, char* desc, int level) : level(level), name(new char[strlen(name) + 1]), desc(new char[strlen(desc) + 1])
{
strcpy_s(this->name, (strlen(name) + 1), name);
strcpy_s(this->desc, (strlen(desc) + 1), desc);
}
Skill::Skill(const Skill& aSkill)
{
this->name = new char[strlen(aSkill.name) + 1];
strcpy_s(this->name, (strlen(aSkill.name) + 1), aSkill.name);
this->level = aSkill.level;
this->desc = new char[strlen(aSkill.desc) + 1];
strcpy_s(this->desc, (strlen(aSkill.desc) + 1), aSkill.desc);
}
Skill& Skill::operator=(const Skill& aSkill)
{
if (this == &aSkill)
return *this;
else
{
delete[] name;
delete[] desc;
name = new char[strlen(aSkill.name) + 1];
strcpy_s(name, (strlen(aSkill.name) + 1), aSkill.name);
desc = new char[strlen(aSkill.desc) + 1];
strcpy_s(name, (strlen(aSkill.desc) + 1), aSkill.desc);
level = aSkill.level;
return *this;
}
}
Skill::~Skill()
{
delete[] name;
delete[] desc;
}
char* Skill::GetName() const
{
return name;
}
char* Skill::GetDesc() const
{
return desc;
}
int Skill::GetLevel() const
{
return level;
}
void Skill::Display(ostream& out)
{
out << "- " << GetName() << " -- " << GetDesc() << " [Lvl: " << GetLevel() << "]\n";
}
Node:
Skill aSkill;
Node* parent;
Node* children[CHILD_MAX];
Node() : parent(NULL)
{
for (int i = 0; i < CHILD_MAX; i++)
{
children[i] = NULL;
}
};
Node(const Skill& n) : aSkill(n), parent(NULL)
{
for (int i = 0; i < CHILD_MAX; i++)
{
children[i] = NULL;
}
};
Here is an extract from main()
SkillTree student("Student");
student.Display(cout);
student.AddSkill("Alphabet","Mastery of letters and sounds",0);
student.Display(cout);
student.AddSkill("Reading","The ability to read all manner of written material",1,"Alphabet");
student.AddSkill("Writing","The ability to put your thoughts on paper",1,"Alphabet");
student.Display(cout);
student.AddSkill("Speed Reading Level 1","Read any text twice as fast as normal",5,"Reading");
student.AddSkill("Speed Reading Level 2","Read any text four times as fast as normal",10,"Speed Reading Level 1");
student.AddSkill("Memorization","Memorize average sized texts",10,"Reading");
student.AddSkill("Massive Memorization","Memorize large sized texts",20,"Memorization");
student.AddSkill("Spell Writing","The ability to write spells",5,"Writing");
student.AddSkill("History","The ability to write (and rewrite) history",10,"Writing");
student.AddSkill("Written Creation","The ability to write things into reality",20,"History");
student.Display(cout);
The two functions that student.Display(cout); calls are as follow
void Tree::Display(ostream& out)
{
out << "Skill Tree: " << title << "\n";
if (this->root == NULL)
{
cout << "Empty\n";
return;
}
else
Display_r(out, this->root, 1);
}
void Tree::Display_r(ostream& out, Node* n, int depth)
{
for (int i = 0; i<depth; i++)
{
out << " ";
}
n->aSkill.Display(out);
for (int i = 0; i<CHILD_MAX; i++)
{
if (n->children[i] != NULL)
{
Display_r(out, n->children[i], depth + 1);
}
}
}
If I comment out a section of code in the Queue implementation of AddSkill(), I get no error.

In the first AddSkill() you insert the new node on the top of the tree, making it the new root.
In the second AddSkill() you intend to insert the new node as child of a parent skill. The approach seems to be:
check that there is at least one node in the tree (initial if)
traverse the tree to find the parrent node ( while loop )
if the parent is found, find the first empty child to insert the new skill (inner for loop)
What are the problems ?
THere are several flaws in your algorithm:
you loop on root not null. As the tree is not empty here, and as you don't delete any node, this condition will remain true, allowing for an endless loop.
then you check if the new child's name corresponds to the parentname. I assume that this will be false most of the case (otherwhise you'd need one parameter less). So this will ensure that the loop is endless.
later you assume that node is the current node, and you insert the new child into node's children. This code is not exectuted. Fortunately: it would be undefined behaviour, because you've set node to NULL and never changed this value.
How to solve it ?
To do this right, you'd have to start with node at root, then check if the node's name matches parentname, and if yes, insert the child as you did.
There's a last problem however. A rather important one. The structure of your algorithm works for a linked list traversal, but not a tree traversal. The tree traversal algorithm requires either a stack/list to keep track of all the branches to explore, or a recursive approach.
Here some code (sorry, I've replaced char* with string and used vector<Node*> instead of Node*[]), using an auxiliary overload of AddSkill(), to perform the recursive search:
// replaces the former one that you had
void Tree::AddSkill(string name, string desc, int level, string parentName)
{
if (root == NULL)
{
cout << "Error: no nodes in tree.\n";
return;
}
Skill s(name, desc, level);
AddSkill(root, s, parentName);
}
// auxiliary helper
void Tree::AddSkill(Node*node, Skill& s, string& parentName)
{
if (node->sk.name == parentName) { // if found, add the new node as childen
Node * child = new Node(s);
child->parent = node;
node->children.push_back(child);
}
else {
for (auto &x : node->children) // for all the children
AddSkill(x, s, parentName); // search recursively
}
}
And here an online demo using shared pointers instead of raw pointers.

Related

basic insertion sort in linked list with logic problem (maybe it is because pointer)

I write a code for insertion sort for integer data in linked list in c++, I referred to the algorithms on the Internet, and finally took the following code using array as a basic concept for my version.
however, the sorting always ignore my first element,(but all the other element is ordered well).
I have tried checking my loop statement, checking the pointer address while looping (because my key pointer loop at first time didn't go into the judge pointer loop), checking the shifting mechanism while comparing, but I cannot find my logic problem.
(I know someone would said I doesn't provide enough data for you to help me, but I have been checking these things for two days, including asking friends and searching the solutions existed on website. So I really hope someone can answer me without blame, thank you.)
array version(on the internet)
#include <iostream>
void InsertionSort(int *arr, int size){
for (int i = 1; i < size; i++) {
int key = arr[i];
int j = i - 1;
while (key < arr[j] && j >= 0) {
arr[j+1] = arr[j];
j--;
}
arr[j+1] = key;
}
}
linked list version(by my own)
Node class used in my version
class Node
{
public:
Node()
{
next = NULL;
pre = NULL;
}
Node(int n)
{
data = n;
next = NULL;
pre = NULL;
}
int getData() { return data; }
Node *getNext() { return next; }
Node *getPre() { return pre; }
void setData(int d) { data = d; }
void setNext(Node *n) { next = n; }
void setPre(Node *p) { pre = p; }
private:
int data;
Node *next, *pre;
};
class List
{
public:
List() { list = NULL; }
List(int n) { generate(n); }
void generate(int n)
{
int j;
list = NULL;
for(j = 0;j < n;j ++)
generate();
}
void generate()
{
Node *buf = new Node(rand());
buf->setNext(list); //list->NODE2.next->NODE1.next->NULL
if(list != NULL)
list->setPre(buf);
list = buf;
}
void insertionSort()
{
bool breakByCompare;
Node* keyptr;
Node* judgeptr;// judge is the value that is going to compare with key
int key;
for(keyptr = list->getNext(); keyptr != NULL;
keyptr = keyptr->getNext()){
//if we set list as 5,7,6 ; 6 is key
key = keyptr->getData();//store the key value for the setting after shifting
breakByCompare = 0;
for(judgeptr = keyptr->getPre() ; judgeptr->getPre()!= NULL;
judgeptr= judgeptr->getPre()){
//list: 5,7,6 ; 7 is judge
if(judgeptr->getData() > key){
// 7>6, so we shift 7 to the position which was for 6
judgeptr->getNext()->setData(judgeptr->getData());// list: 5,7,7 ;
cout << judgeptr->getData() << " , " << keyptr->getData() << endl;
}
else{
break;
}
}
judgeptr->getNext()->setData(key);// list: 5,6,7
}
}
void print()
{
Node *cur = list;
while(cur != NULL)
{
cout<<cur->getData()<<" ";
cur = cur->getNext();
}
cout<<endl;
}
private:
Node *list;
};
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <ctime>
#define SIZE 100
int main()
{
srand(time(NULL));
List *l = new List(10);
l->print();
l->insertionSort();
l->print();
}
One of the most important difference between a linked list and an array is that it is sometimes required to handle the first element as a special case.
Here is a fixed version of your sorting method :
void insertionSort()
{
bool breakByCompare;
Node* keyptr;
Node* judgeptr;
int key;
for(keyptr = list->getNext(); keyptr != NULL; keyptr = keyptr->getNext()){
key = keyptr->getData();
breakByCompare = 0;
// I replaced judgeptr->getPre() by judgeptr in the condition
// to allow the backward loop to go until the root
for(judgeptr = keyptr->getPre() ; judgeptr != NULL; judgeptr= judgeptr->getPre()){
if(judgeptr->getData() > key){
judgeptr->getNext()->setData(judgeptr->getData());
cout << judgeptr->getData() << " , " << key << endl;
}
else break;
}
// Here is the special case : we must support a null judgeptr
// and replace its next element by the list
if (judgeptr) judgeptr->getNext()->setData(key);
else list->setData(key);
}
}

Function couldn't been resolved

I have a problem with my C++ code. It says that the all the functions starting with isPerfectRec() couldn't be resolved...Why? I tried a lot of things but apparently they don't work. I have a lot of assigments like to verify if the binary search tree is perfect, to find the second largest element in a binary search tree and so on..
#include <stdio.h>
#include<iostream>
#include<stack>
template<typename T> class BinarySearchTree {
public:
BinarySearchTree<T> *root, *left_son, *right_son, *parent;
T *pinfo;
BinarySearchTree() {
left_son = right_son = NULL;
root = this;
pinfo = NULL;
}
void setInfo(T info) {
pinfo = new T;
*pinfo = info;
}
void insert(T x) {
if (pinfo == NULL)
setInfo(x);
else
insert_rec(x);
}
bool isPerfectRec(BinarySearchTree *root, int d, int level = 0)
{
// An empty tree is perfect
if (*root == NULL)
return true;
// If leaf node, then its depth must be same as
// depth of all other leaves.
if (*root->left_son == NULL && root->*right_son == NULL)
return (d == level+1);
// If internal node and one child is empty
if (root->*left_son == NULL || root->*right_son == NULL)
return false;
// Left and right subtrees must be perfect.
return isPerfectRec(root->*left_son, d, level+1) &&
isPerfectRec(root->*right_son, d, level+1);
}
// Wrapper over isPerfectRec()
bool isPerfect(BinarySearchTree *root)
{
int d = findADepth(root);
return isPerfectRec(root, d);
}
int findADepth(BinarySearchTree *node)
{
int d = 0;
while (node != NULL)
{
d++;
node = node->left_son;
}
return d;
}
// A function to find 2nd largest element in a given tree.
void secondLargestUtil(BinarySearchTree *root, int &c)
{
// Base cases, the second condition is important to
// avoid unnecessary recursive calls
if (root == NULL || c >= 2)
return;
// Follow reverse inorder traversal so that the
// largest element is visited first
secondLargestUtil(root->right_son, c);
// Increment count of visited nodes
c++;
// If c becomes k now, then this is the 2nd largest
if (c == 2)
{
std::cout << "2nd largest element is "
<< root->pinfo;
printf("\n___\n");
return;
}
// Recur for left subtree
secondLargestUtil(root->left_son, c);
}
void secondLargest(BinarySearchTree *root)
{
// Initialize count of nodes visited as 0
int c = 0;
// Note that c is passed by reference
secondLargestUtil(root, c);
}
bool hasOnlyOneChild(int pre[], int size)
{
int nextDiff, lastDiff;
for (int i=0; i<size-1; i++)
{
nextDiff = pre[i] - pre[i+1];
lastDiff = pre[i] - pre[size-1];
if (nextDiff*lastDiff < 0)
return false;;
}
return true;
}
BinarySearchTree * readListInter(){
BinarySearchTree* root = NULL;//returning object
BinarySearchTree* temp;
BinarySearchTree* input;//new node to add
int x;
std::cout << "enter number (>0 to stop): ";
std::cin >> x;
while(x>=0){
input = BinarySearchTree(x);
if(root == NULL){//if root is empty
root = input;
temp = root;//temp is use to store value for compare
}
else{
temp = root; //for each new addition, must start at root to find correct spot
while(input != NULL){
if( x < temp->pinfo){//if smaller x to add to left
if(temp->left_son == NULL){//left is empty
temp->left_son = input;
input = NULL;//new node added, exit the loop
}
else{//if not empty set temp to subtree
temp = temp->left_son;//need to move left from the current position
}
}
else{//otherwise x add to right
if(temp->right_son == NULL){//right is empty
temp->right_son = input;
input = NULL;//new node added, exit the loop
}
else{
temp = temp->right_son;//need to move right from the current position
}
}
}
}
std::cin >> x;
}
return root;
}
};
int main() {
BinarySearchTree<int> *r = new BinarySearchTree<int>;
BinarySearchTree<int> *r1 = new BinarySearchTree<int>;
BinarySearchTree<int> *p = new BinarySearchTree<int>;
p = readListInter();
r->insert(6);
r->insert(8);
r->insert(1);
r->insert(9);
r->insert(10);
r->insert(4);
r->insert(13);
r->insert(12);
printf("\n___\n");
r1->insert(6);
r1->insert(8);
r1->insert(1);
r1->insert(9);
r1->insert(10);
r1->insert(4);
r1->insert(13);
r1->insert(12);
printf("\n___\n");
r->isPerfect(r);
int pre[] = {8, 3, 5, 7, 6};
int size = sizeof(pre)/sizeof(pre[0]);
if (hasOnlyOneChild(pre, size) == true )
printf("Yes");
else
printf("No");
s
return 0;
}
I think you need to write BinarySearchTree<T> instead of BinarySearchTree as a datatype in those functions.

wanted to return root of tree node with latest addition of a node intact

The issue: the root node is being set to null and the actual addition of nodes in the tree doesn't quite happen:
Client program:
int main () {
int arr [] = {15, 10, 100, 5, 13, 90, 80, 50, 10, 5, 3};
int len = sizeof(arr)/sizeof(*arr);
using namespace btree;
//create an instance of btree::tree;
btree::tree* tree = new btree::tree();
btree::btnode* root = tree->get_root_btnode();
//create tree
tree->create_btree (root, arr, len);
delete tree;
return 0;
}
The member functions (used in client program) looks like follows:
//creates a complete tree
void tree::create_btree (btnode* root, int arr[], int len) {
for (int i = 0; i < len; i++) {
add_a_btnode (root, arr[i]);
}
std::cout << "Tree creation successful!!" << std::endl;
}
void tree::add_a_btnode (btnode* root, int data) {
///check if tree is empty
btnode *current = root;
if (!current) {
btnode* newbtnode = new btnode(data);
root = newbtnode;
return;
}
if (current->get_left_btnode() != nullptr) {
//can we add it as right child of current btnode?
if (current->get_right_btnode() != nullptr) {
//recurse
add_a_btnode (current->get_left_btnode(), current->get_data());
} else {
btnode* newbtnode = new btnode(data);
current->m_right = newbtnode;
}
} else {
btnode* newbtnode = new btnode(data);
current->m_left = newbtnode;
}
std::cout << "a root btnode is added with data: " << data << endl;
}
and in last the member functions which i expected to return me the root node (which doesn't quite happen :( any help):
btnode* tree::get_root_btnode () { return m_root_btnode; }
Your problem is here:
if (!current) {
btnode* newbtnode = new btnode(data);
root = newbtnode;
return;
}
You create root node and save it to variable root. But it is a local variable, and in fact you don't modify the tree object at all.
Instead, you should save it to variable inside your tree class. So it would be something like that:
if (!current) {
m_root_btnode = new btnode(data);
return;
}
Btw. Why do you pass pointer to root as argument, when the functions are methods of tree class and you have access to original root variable? This would be much cleaner, and you wouldn't make that mistake then.
Change the create_btree function as follows:
void tree::create_btree(int arr[], int len) {
if (len > 0) {
m_root_btnode = new btnode(arr[0]);
}
for (int i = 1; i < len; i++) {
add_a_btnode(m_root_btnode, arr[i]);
}
std::cout << "Tree creation successful!!" << std::endl;
}
There are many issues with your code, but this is what I'd start with.

Recursively Populate N-ary Tree in c++

I'm attempting to create the game tree of the tic-tac-toe board game. I've written some basic methods, but I'm having trouble recursively populating the elements of the tree.
I'm using a Node struct to define the Nodes of the tree. Each node has an array of children.
struct node {
string data;
int height;
node * child[9];
};
Each Node stores the content of a game board as a string. * is used to display blanks.
So, * * * * * * * * * would be a blank board.
I have a Tree class that implements the tree.
class Tree {
public:
Tree();
Tree(string data);
~Tree();
void insert(string data, node * leaf);
node * get_root();
void populate(node * n);
void generate_tree(node * n);
int number_of_blanks(string);
private:
void destroy_tree(node * leaf);
node * root;
node * temp;
int count;
};
Tree::Tree(string data) {
root = new node;
root->data = data;
root->height = 0;
temp = root;
count = 0;
}
Here is my method for inserting nodes. It inserts a new node to the first NULL child.
void Tree::insert(string data, node * leaf) {
int i;
//checks for first NULL child
for(i = 0; i < 9; i++) {
if(leaf->child[i] == NULL) {
//first NULL child is inserted and all its children set to NULL
leaf->child[i] = new node;
leaf->child[i]->data = data;
leaf->child[i]->height = leaf->height + 1;
break;
}
}
}
This code works how I've intended it to, however I'm sure it's not the best method.
Where I'm having the most trouble is recursively populating the tree. My recursion either ends early, or is an endless loop. I'm not sure how to approach this problem, as I've never used recursion with a void method.
void Tree::generate_tree(node * leaf) {
int i;
string data;
string player;
int length = number_of_blanks(leaf->data);
if(leaf->height % 2 == 0)
player = "X";
else
player = "O";
if(leaf->data.find_last_of('*',8) == string::npos) {
cout << "This is a leaf!!!!!!!!! " << leaf->data << endl;
return;
}
for(i = 0; i < length; i++) {
if(leaf->height >=9 )
return;
data = leaf->data.replace(count,1,player);
insert(data,leaf);
cout << "New Node: " << data << " Height: " << leaf->child[i]->height << endl;
count++;
generate_tree(leaf->child[i]);
count = 0;
}
}
Any tips, or specific suggestions would be greatly appreciated. Thank you.
I'd recommend giving node a constructor and to initialize members before the code block that makes up the constructor. For example:
node(string s, int h) : data(s), height(h) {
for (int i=0;i < 9; ++i)
child[i] = NULL;
}
Similarly for the constructor for Tree:
Tree::Tree(std::string data) : root(new node(data,0)), count(0) {}
This makes other parts of the code much simpler. For example, your insert code would now look like this:
void Tree::insert(std::string data, node * leaf) {
//checks for first NULL child
for(int i = 0; i < 9; i++) {
if(leaf->child[i] == NULL) {
//first NULL child is inserted and all its children set to NULL
leaf->child[i] = new node(data, leaf->height+1);
break;
}
}
}
I haven't had the time to analyze the rest, but this may make it easier for you to solve.

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);
}