Logical error in insertion of node in a BST - c++

I have written a node insertion code for a BST. But it dosen't seem to work correctly. Its giving a "Segmentation error". Here is my logic for insertion.
void insert(Node* root, int data){
if(root==NULL){
root= new Node;
root->data = data;
}
else if(data < root->data){
insert(root->left,data);
}
else if(data> root->data){
insert(root->right,data);
}
}
How do i fix it? Thanks
Edit: so i tried out some things and this one does the trick
Node* insert(Node* &root, int data){
if(root==nullptr){
root = create(data);
return root;
}
else if(data < root->data){
insert(root->left,data);
}
else if(data> root->data){
insert(root->right,data);
}
}
Whats the difference between Node* root and Node* &root ?

Well, if node doesn't exists (it's NULL), you're just setting your root pointer to new Node, but you're missing to 'hang it up' to it's parent. And as already mentioned, you can use unique_ptr-s since C++11 to avoid memory leaks (that's when you forget to delete object). It looks like:
struct Node {
int data = -1; // not initialized
std::unique_ptr<Node> l;
std::unique_ptr<Node> r;
}
void insert(Node *root, int data) {
if (root->data == -1) {
root->data = data;
return;
}
if (data < root->data) {
if (!root->l) {
// hanging new left node up
root->l = std::make_unique<Node>(); // std::make_unique comes since C++14
}
insert(root->l.get(), // getting raw ptr
data);
}
else {
if (!root->r) {
// hanging new right node up
root->r = std::make_unique<Node>();
}
insert(root->r.get(), data);
}
}
Also you might be interested in data structure called treap, because your implementation may work very long if you insert, for example, increasing sequence:
Node root;
for (int i = 1; i <= 100'000; i++) {
insert(&root, i);
}
So your binary tree in this case looks like:
1
\
2 <=
\ <= very long path
3 <=
\
...
\
100'000
Treap helps to avoid long paths in your BST.

Related

Implementation of BST C++ Segmentation Fault

I have implemented binary search tree in C++ and for some reason I am not seeing where the segmentation fault occurs. But I do notice that when I comment out root = node in the first conditional statement in addNode the error goes away. What exactly is a segmentation fault and how does it related to pointers?
#include <iostream>
#include <iomanip>
using namespace std;
class bstNode
{
public:
int value;
bstNode *left;
bstNode *right;
bstNode(){};
~bstNode(){};
bstNode(int value)
{
this->value = value;
this->left = NULL;
this->right = NULL;
}
bstNode(int value, bstNode *left, bstNode *right)
{
this->value = value;
this->left = left;
this->right = right;
}
bstNode *root;
void addNode(int value)
{
if (root == NULL)
{
bstNode *node = new bstNode(value);
root = node;
}
else
{
bstNode *focusNode = root;
bstNode *parent;
while (focusNode != NULL)
{
if (value > focusNode->value)
{
focusNode = focusNode->right;
if (focusNode == NULL)
{
focusNode->right = new bstNode(value);
}
}
else
{
focusNode = focusNode->left;
if (focusNode == NULL)
{
focusNode->left = new bstNode(value);
}
}
}
}
}
static void printBST(bstNode *node)
{
while (node != NULL)
{
printBST(node->left);
cout << node->value;
printBST(node->right);
}
}
};
int main()
{
bstNode *node = new bstNode();
node->addNode(7);
node->addNode(2);
node->addNode(18);
node->addNode(6);
node->addNode(4);
node->addNode(23);
bstNode::printBST(node->root);
return 0;
}
The immediate error is this
if (focusNode == NULL) {
focusNode->left = new bstNode(value);
}
this is clearly wrong, if a pointer is null you cannot use it. You have this in multiple places. Fix that and then update the question once you have got past that. How did I know this? I ran your code under my debugger and it told me immediatley, you should learn how to get the most out of your debugger.
Next
void addNode(int value)
as a method for a class defined as
class bstNode {
public:
int value;
is very bad practice. In that method what does value refer to? The argument or the member variable. Get into the habit of giving member variables specific names like this
class bstNode {
public:
int value_;
Also minor nits. The accepted style for naming classes is with leading Caps like this
class BstNode {
public:
int value_;
or even
class BSTNode
class bstNode {
public:
int value_;
using namespace std;
I'd advise against doing this in general. It's hard to be sure what's in namespace std, but the short summary is "a lot, and more all the time", so making all of it visible directly can lead to problems.
bstNode(){};
~bstNode(){};
These don't really accomplish anything useful. The point of a constructor is to initialize the object, but these just leave the object uninitialized, which can lead to problems--especially segmentation faults when/if you try to dereference an uninitialized pointer.
bstNode(int value){
this->value = value;
this->left = NULL;
this->right = NULL;
}
This is better, but I'd prefer to use a member initializer list instead of assignments inside the body of the ctor, and I'd prefer nullptr over NULL:
bstNode(int value)
: value(value)
, left(nullptr)
, right(nullptr) {}
This next one:
bstNode(int value, bstNode* left, bstNode* right){
this->value = value;
this->left = left;
this->right = right;
}
...is pretty nicely written (though it could also use a member initializer list, which is usually preferable), but only rarely useful when building a binary search tree, because in normal use you only ever insert new leaf nodes, not new internal nodes.
void addNode(int value){
if (root == NULL){
bstNode* node = new bstNode(value);
root = node;
}
else{
bstNode* focusNode = root;
bstNode* parent;
while(focusNode != NULL){
if(value > focusNode->value){
focusNode = focusNode->right;
if(focusNode == NULL){
focusNode->right = new bstNode(value);
}
}
else{
focusNode = focusNode->left;
if(focusNode == NULL){
focusNode->left = new bstNode(value);
}
}
}
}
}
This is at least one obvious source of a segmentation fault--you dereference a pointer immediately after verifying that it's null.
At least for a first attempt, I think I'd use a recursive implementation, which tends to be simpler:
void addNode(int value, bstNode *&node = root) {
if (node == nullptr) {
node = new node(value);
} else if (value < node->value) {
addNode(value, node->left);
} else if (value > node->value) {
addNode(value, node->right);
} else {
// attempt at inserting duplicate value
}
}
Note that this passes a reference to a pointer, so we can modify the "current" pointer, rather than having to track the parent pointer while traversing the tree.
static void printBST(bstNode* node){
while(node != NULL){
printBST(node->left);
cout << node->value;
printBST(node->right);
}
}
Since we're doing this recursively, we don't need (or even want) a loop. Traversing the left sub-tree, the current node, and the right subtree traverses the entire tree, with no iteration needed.
Also note that this doesn't print any delimiter between the numbers in the nodes, so a tree containing 12, 34 and a tree containing 1, 2, 3, 4 will both be printed out as 1234, which probably isn't very useful. Fortunately, adding a delimiter is pretty easy.
static void printBST(bstNode* node){
if (node != nullptr){
printBST(node->left);
cout << node->value << ' ';
printBST(node->right);
}
}
In the the following code...
while(focusNode != NULL){
if(value > focusNode->value){
focusNode = focusNode->right;
if(focusNode == NULL){
focusNode->right = new bstNode(value);
}
}
else{
focusNode = focusNode->left;
if(focusNode == NULL){
focusNode->left = new bstNode(value);
}
}
}
...you are referencing the children of a node that is guaranteed to be NULL because you verified that using the conditional statement. Since the node itself does not exist, it doesn't have properties like children. Imagine you're trying to communicate with the child of a person who has never existed.
The variable focusNode stores an address of a node. What focusNode->value does is that it goes to the node whose address focusNode stores and retrieves the value property from there.
When focusNode is NULL, it doesn't point to any node, thus you can't go there and retrieve its value property.
I wrote the code that you can replace with your while loop. I have tested it and it works:
while(true){
if(value > focusNode->value){
if(focusNode->right == NULL){
focusNode->right = new bstNode(value);
return;
} else focusNode = focusNode->right;
}
else{
if(focusNode->left == NULL){
focusNode->left = new bstNode(value);
return;
} else focusNode = focusNode->left;
}
}
I also fixed your printBST function. In the printBST function use if instead of while, because the the code inside the while loop would be executed an infinite number of times instead of printing the BST once.
static void printBST(bstNode* node){
if(node != NULL){
printBST(node->left);
cout << node->value <<" ";
printBST(node->right);
}
}

I wanted to implement a BST and tried using vector for input

I wanted to implement a BST class with a vector and somehow its not working. I just wanted to know the reason why its not working.
The main reason that I can think of that root in the BST always remain NULL.
I wanted to experiment ways to use classes in data structures.
#include<iostream>
#include<vector>
using namespace std;
class Node{
public:
int data;
Node* left ;
Node* right ;
Node(int val){
data = val;
left = NULL;
right = NULL;
}
};
class BST{
public:
Node* root = NULL;
void insert(Node* r,int data){
Node* new_node = new Node(data);
if(r == NULL){
r = new_node;
}
if(data < r->data){
if(r->left == NULL){
r->left = new_node;
}
else{
insert(r->left,data);
}
}else if(data > r->data){
if(r->right == NULL){
r->right = new_node;
}
else{
insert(r->right,data);
}
}else{
return;
}
return;
}
BST(vector<int> bst_array){
for(int i = 0; i<bst_array.size(); i++){
insert(root,bst_array[i]);
}
}
void print_t(Node* r){
if(r == NULL){
cout<<"NULL";
return;
}
else{
print_t(r->left);
cout<<r->data<<" ";
print_t(r->right);
}
}
};
int main(){
vector<int> v = {1,3,5,44,23,78,21};
BST* tr = new BST(v);
tr->print_t(tr->root);
return 0;
}
There seem to be a logical mistake on my end please help me find it.
Thanks in advance.
The reason is that root is never assigned another value after its initialisation to NULL. Passing root as argument to the insert method can never alter root itself, as it is not the address of root that is passed, but its value.
Some other remarks:
insert always starts by creating a new node, at every step of the recursion. This is a waste of node creation. In the end you just need one new node, so only create it when its position in the tree has been identified.
The final else is not needed, as all it does is execute a return, which it would have done anyway without that else block
As insert is a method of BST, it is a pity that it requires a node as argument. You would really like to just do insert(data) and let it take care of it. For that to happen I suggest to move your insert method to the Node class, where the this node takes over the role of the argument. Then the BST class could get a wrapping insert method that forwards the job to the other insert method.
Instead of NULL use nullptr.
To solve the main issue, there are many solutions possible. But after making the above changes, it is quite easy to assign to root in the simplified insert method on the BST class.
Here is how it could work:
class Node{
public:
int data;
Node* left ;
Node* right ;
Node(int val){
data = val;
left = nullptr;
right = nullptr;
}
void insert(int data) {
if (data < this->data) {
if (this->left == nullptr) {
this->left = new Node(data);
} else {
this->left->insert(data);
}
} else if (data > this->data) {
if (this->right == nullptr) {
this->right = new Node(data);
} else {
this->right->insert(data);
}
}
}
};
class BST {
public:
Node* root = nullptr;
void insert(int data) {
if (root == NULL) { // Assign to root
root = new Node(data);
} else { // Defer the task to the Node class
root->insert(data);
}
}
BST(vector<int> bst_array){
for(int i = 0; i<bst_array.size(); i++){
insert(bst_array[i]); // No node argument
}
}
/* ...other methods ...*/
}

Inserting elements in a bst using recursion

I'm trying to add elements enter by the user in a BST.For this I've used 2 functions, one is used to create the function and other is just used to insert element to the tree. One is a pre-order function that is used to check if insertion is done or not Initially I tried to add elements manually.Its not printing all inserted values.
The overall layout
struct Node{
int data;
struct Node* left;
struct Node* right;
};
void Inorder(struct Node* root){
if(root==NULL){
return;
}
else{
Inorder(root->left);
cout<<root->data<<" ";
Inorder(root->right);
}
}
struct Node* create_node(int data){
struct Node* node=(struct Node*) malloc(sizeof(struct Node));
node->data=data;
node->left=NULL;
node->right=NULL;
return node;
}
The problem code:-
struct Node* insert(struct Node* root,int data){
static struct Node* prev=NULL;
if(root==NULL && prev==NULL){
return create_node(data);
}
if(root->data==data){
return root;
}
else{
if(root==NULL){
struct Node* ptr=create_node(data);
if(prev->data>data){
prev->left=ptr;
return root;
}
else{
prev->right=ptr;
return root;
}
}
else{
if(root->data>data){
prev=root;
insert(root->left,data);
}
else{
prev=root;
insert(root->right,data);
}
}
}
}
MAIN
int main()
{
struct Node* root=NULL;
root=insert(root,5);
Inorder(root);
cout<<endl;
insert(root,3);
Inorder(root);
insert(root,10);
Inorder(root);
return 0;
}
One thing I noticed that prev is static once we call insert for inserting next element(here 3) it won't roll over from start again because it is declared static.To overcome that
Tried to optimize the problem code by making prev as global and making null in main every time I call insert function in the main(), The optimised code is as follows:
#include <iostream>
#include<stdlib.h>
using namespace std;
static struct Node* prev=NULL;
struct Node{
int data;
struct Node* left;
struct Node* right;
};
void Inorder(struct Node* root){
if(root==NULL){
return;
}
else{
Inorder(root->left);
cout<<root->data<<" ";
Inorder(root->right);
}
}
struct Node* create_node(int data){
struct Node* node=(struct Node*) malloc(sizeof(struct Node));
node->data=data;
node->left=NULL;
node->right=NULL;
return node;
}
struct Node* insert(struct Node* root,int data){
if(root==NULL && ::prev==NULL){
return create_node(data);
}
if(root->data==data){
return root;
}
else{
if(root==NULL){
struct Node* ptr=create_node(data);
if(::prev->data>data){
::prev->left=ptr;
return root;
}
else{
::prev->right=ptr;
return root;
}
}
else{
if(root->data>data){
::prev=root;
insert(root->left,data);
}
else{
::prev=root;
insert(root->right,data);
}
}
}
}
int main()
{
struct Node* root=NULL;
root=insert(root,5);
Inorder(root);
cout<<endl;
::prev=NULL;
insert(root,3);
Inorder(root);
::prev=NULL;
insert(root,10);
Inorder(root);
return 0;
}
This is not how insertion into a BST is supposed to work. You don't need a prev pointer at all.
One of the issues in your code is that you don't use the return value of the recursive call, which at some point is going to be the pointer to a new node! You should really assign that return value to either the left or right member of the current node.
Also, the following if condition will never be true, as at that point it was already guaranteed that root is not NULL:
else{
if(root==NULL){
The correct code is actually quite simple:
struct Node* insert(struct Node* root, int data){
if (root == NULL) {
root = create_node(data);
} else if (root->data > data) {
root->left = insert(root->left, data);
} else if (root->data < data) {
root->right = insert(root->right, data);
}
return root;
}
I would also add some line breaks to the output in your main code:
int main()
{
struct Node* root = NULL;
root = insert(root, 5);
Inorder(root);
cout << endl;
insert(root, 3);
Inorder(root);
cout << endl;
insert(root, 10);
Inorder(root);
cout << endl;
return 0;
}
The issue that I noticed in my code (Unfortunately unable to upload the snippet). was in the part mentioned below.
if(root->data==data){
return root;
}
Firstly let me explain the recursion function, at beginning the root would be null at insertion(here inserting 5 as root) so first condition will be satisfied i.e.
if(root==NULL && ::prev==NULL){
return create_node(data);
}
and the function would return,now I set the global variable prev as NULL because I want to traverse again from the root of the tree to add the next element.
Now once we try to add another element (here adding element 3). This condition
if(root==NULL && ::prev==NULL){
return create_node(data);
}
won't be true, now the thought process while writing the logic was checking if at some stage while traversing down the tree if we encounter node with same value then we'll return the root and terminate the function. This is what I tried to implement .
Here's the code if you could relate(Problem Code Snippet)
else if(root->data==data){
return root;
}
No doubt approach is fine but I forgot to add one condition(actually I preempted that at this stage the root won't be NULL) but root can be NULL.
Because of this we will face segmentation fault error (in debugger mode -> which helped me to find the error in my code!).
So the correct code would be:
else if(root && root->data==data){// or if(root!=NULL && root->data=data)
return root;
}
Rest of the code remains unaltered
So to sum up when traversing through tree we return true for all conditions and once we reach NULL then since first condition won't we satisfied as prev!=NULL, so it comes to next condition root->data==data but here root=NULL so we get
segmentation fault error and function never encounters ROOT==NULL which was designed for this purpose only i.e. to add/insert element in the tree as everything seems fine on traversing the tree. So to over come this problem I modified my else if condition i.e. else if(root && root->data==data)
so the full function code is as follows:
struct Node* insert(struct Node* root,int data){
if(root==NULL && ::prev==NULL){
return create_node(data);
}
else if(root && root->data==data){
return root;
}
else{
if(root==NULL){
struct Node* ptr=create_node(data);
if(::prev->data>data){
::prev->left=ptr;
return root;
}
else{
::prev->right=ptr;
return root;
}
}
else{
if(root->data>data){
::prev=root;
insert(root->left,data);
}
else{
::prev=root;
insert(root->right,data);
}
}
}
}
PS: The code was executed for many trees including one mentioned in the question and got the expected results i.e. Inorder was a sorted array which depicts that insertion was done correctly.

Binary Tree Insert Algorithm

I recently finished implementing a Binary search tree for a project I was working on. It went well and I learned a lot. However, now I need to implement a regular Binary Tree... which for some reason has me stumped.
I'm looking for a way to do my InsertNode function..
normally in a BST you just check if data < root then insert left and vice versa. However, In a normal Binary tree, it is just filled from left to right, one level at a time..
could anyone help me implement a function that just adds a new Node to the Binary tree from left to right in no specific order?
Here's my Insert for a BST:
void Insert(Node *& root, int data)
{
if(root == nullptr)
{
Node * NN = new Node;
root = NN;
}
else
{
if(data < root->data)
{
Insert(root->left, data);
}
else
{
Insert(root->right, data);
}
}
}
I am aware of the fact that this is a question posted some time ago, but I still wanted to share my thoughts on it.
What I would do (since this indeed is not very well documented) is use a Breadth-First-Search (using a queue) and insert the child into the first null I encounter. This will ensure that your tree will fill up the levels first before it goes to another level. With the right number of nodes, it will always be complete.
I haven't worked that much with c++, so to be sure it was correct I did it in Java, but you get the idea:
public void insert(Node node) {
if(root == null) {
root = node;
return;
}
/* insert using Breadth-first-search (queue to the rescue!) */
Queue<Node> queue = new LinkedList<Node>();
queue.offer(root);
while(true) {
Node n = queue.remove();
if(!n.visited) System.out.println(n.data);
n.visited = true;
if(n.left == null) {
n.left = node;
break;
} else {
queue.offer(n.left);
}
if(n.right == null) {
n.right = node;
break;
} else {
queue.offer(n.right);
}
}
}
Javascript implementation (copy-paste ready for your web console):
ES6 implementation (newer javscript syntax with class keyword)
class BinaryTree {
constructor(value){
this.root = value;
this.left = null;
this.right = null;
}
insert(value){
var queue = [];
queue.push(this); //push the root
while(true){
var node = queue.pop();
if(node.left === null){
node.left = new BinaryTree(value);
return;
} else {
queue.unshift(node.left)
}
if(node.right === null){
node.right = new BinaryTree(value);
return;
} else {
queue.unshift(node.right)
}
}
}
}
var myBinaryTree = new BinaryTree(5);
myBinaryTree.insert(4);
myBinaryTree.insert(3);
myBinaryTree.insert(2);
myBinaryTree.insert(1);
5
/ \
4 3
/ \ (next insertions here)
2 1
Pseudoclassical pattern implementation
var BinaryTree = function(value){
this.root = value;
this.left = null;
this.right = null;
}
BinaryTree.prototype.insert = function(value){
//same logic as before
}
With a few modifications to your code, I hope this should help :
Node * Insert(Node * root, int data)
{
if(root == nullptr)
{
Node * NN = new Node();
root = NN;
root->data = data;
root->left = root ->right = NULL;
}
else
{
if(data < root->data)
{
root->left = Insert(root->left, data);
}
else
{
root->right = Insert(root->right, data);
}
}
return root;
}
Hence , this function returns the root node of the updated BST.
I took bknopper code, modified a little bit and translated to C++. As he stated, surprisingly, this is not well documented.
Here is the node structure and the insert function:
struct nodo
{
nodo(): izd(NULL), der(NULL) {};
int val;
struct nodo* izd;
struct nodo* der;
};
void inserta(struct nodo** raiz, int num)
{
if( !(*raiz) )
{
*raiz = new struct nodo;
(*raiz)->val = num;
}
else
{
std::deque<struct nodo*> cola;
cola.push_back( *raiz );
while(true)
{
struct nodo *n = cola.front();
cola.pop_front();
if( !n->izd ) {
n->izd = new struct nodo;
n->izd->val = num;
break;
} else {
cola.push_back(n->izd);
}
if( !n->der ) {
n->der = new struct nodo;
n->der->val = num;
break;
} else {
cola.push_back(n->der);
}
}
}
}
You call it this way:
inserta(&root, val);
Being root a pointer to node struct and val the integer value you want to insert.
Hope it helps someone.
You should try using a recursive approach such as x = new (x), if you know what that means. This way, you don't really have to worry about the root node. I am going to write some pseudocode for you:
//public function
add(data){
root = add(data, root)
}
//private helper function
Node add(data, currentNode){
if(currentNode = 0)
return new Node(data)
if(data less than currentNode's data)
currentNode.left = add(data, currentNode.left)
if(data more than currentNode's data)
currentNode.right = add(data, currentNode.right)
return currentNode
}
I made a tutorial regarding the implementation of a BST in C++, here

BST Insert C++ Help

typedef struct treeNode {
treeNode* left;
treeNode* right;
int data;
treeNode(int d) {
data = d;
left = NULL;
right = NULL;
}
}treeNode;
void insert(treeNode *root, int data) {
if (root == NULL) {
cout << &root;
root = new treeNode(data);
}
else if (data < root->data) {
insert(root->left, data);
}
else {
insert(root->right, data);
}
}
void inorderTraversal(treeNode* root) {
if (root == NULL)
return;
inorderTraversal(root->left);
cout<<root->data;
inorderTraversal(root->right);
}
int main() {
treeNode *root = new treeNode(1);
cout << &root << endl;
insert(root, 2);
inorderTraversal(root);
return 0;
}
So I'm pretty tired, but I was whipping some practice questions up for interview prep and for some reason this BST insert is not printing out that any node was added to the tree. Its probably something im glossing over with the pointers, but I can't figure it out. any ideas?
void insert(treeNode *root, int data) {
if (root == NULL) {
cout << &root;
root = new treeNode(data);
}
This change to root is lost as soon as the function ends, it does not modify the root passed as argument but its own copy of it.
Take note that when u insert the node, use pointer to pointer (pointer alone is not enough):
So, here is the fixed code:
void insert(treeNode **root, int data) {
if (*root == NULL) {
cout << root;
*root = new treeNode(data);
}
else if (data < (*root)->data) {
insert(&(*root)->left, data);
}
else {
insert(&(*root)->right, data);
}
}
And in main:
int main() {
treeNode *root = new treeNode(1);
cout << &root << endl;
insert(&root, 2);
inorderTraversal(root);
return 0;
}
Your logic is correct!
The only issue is that when you create a local variable, even if it is a pointer, its scope is local to the function. In your main:
...
insert(root, 2);
...
function call sends a copy of the root which is a pointer to treeNode (not the address of root). Please note that
void insert(treeNode *root, int data)
gets a treeNode pointer as an argument (not the address of the pointer). Attention: This function call may look like "call by pointer" (or reference) but it is actually "call by value". The root you define in the main function and the root inside the insert method have different addresses in the stack (memory) since they are different variables. The former is in main function stack in the memory while the latter is in insert method. Therefore once the function call insert finishes executing, its stack is emptied including the local variable root. For more details on memory refer to: stacks/heaps.
Of course the data in the memory that you allocated using:
*root = new treeNode(data);
still stays in the heap but you have lost the reference to (address of) it once you are out of the insert function.
The solution is either passing the address of original root to the function and modifying it (as K-ballo and dip has suggested) OR returning the modified local root from the function. For the first approach please refer to the code written by dip in his/her answer.
I personally prefer returning the modified root from the function since I find it more convenient especially when implementing other common BST algorithms. Here is your function with a slight modification of your original code:
treeNode* insert(treeNode *root, int data) {
if (root == NULL) {
root = new treeNode(data);
}
else if (data < root->data) {
root->left=insert(root->left, data);
}
else {
root->right=insert(root->right, data);
}
return treeNode;
}
The function call in main will be:
int main() {
treeNode *root = new treeNode(1);
cout << &root << endl;
root = insert(root, 2);
inorderTraversal(root);
return 0;
}
Hope that helps!
After a while seeing some complicated methods of dealing with the Binary tree i wrote a simple program that can create, insert and search a node i hope it will be usefull
/*-----------------------Tree.h-----------------------*/
#include <iostream>
#include <queue>
struct Node
{
int data;
Node * left;
Node * right;
};
// create a node with input data and return the reference of the node just created
Node* CreateNode(int data);
// insert a node with input data based on the root node as origin
void InsertNode (Node* root, int data);
// search a node with specific data based on the root node as origin
Node* SearchNode(Node* root, int data);
here we define the node structure and the functions mentioned above
/*----------------------Tree.cpp--------------*/
#include "Tree.h"
Node* CreateNode(int _data)
{
Node* node = new Node();
node->data=_data;
node->left=nullptr;
node->right=nullptr;
return node;
}
void InsertNode(Node* root, int _data)
{
// create the node to insert
Node* nodeToInsert = CreateNode(_data);
// we use a queue to go through the tree
std::queue<Node*> q;
q.push(root);
while(!q.empty())
{
Node* temp = q.front();
q.pop();
//left check
if(temp->left==nullptr)
{
temp->left=nodeToInsert;
return;
}
else
{
q.push(temp->left);
}
//right check
if(temp->right==nullptr)
{
temp->right=nodeToInsert;
return;
}
else
{
q.push(temp->right);
}
}
}
Node* SearchNode(Node* root, int _data)
{
if(root==nullptr)
return nullptr;
std::queue<Node*> q;
Node* nodeToFound = nullptr;
q.push(root);
while(!q.empty())
{
Node* temp = q.front();
q.pop();
if(temp->data==_data) nodeToFound = temp;
if(temp->left!=nullptr) q.push(temp->left);
if(temp->right!=nullptr) q.push(temp->right);
}
return nodeToFound;
}
int main()
{
// Node * root = CreateNode(1);
// root->left = CreateNode(2);
// root->left->left = CreateNode(3);
// root->left->left->right = CreateNode(5);
// root->right = CreateNode(4);
// Node * node = new Node();
// node = SearchNode(root,3);
// std::cout<<node->right->data<<std::endl;
return 0;
}