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Iterative Search for a key ‘x’ in Binary Tree

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  • Difficulty Level : Easy
  • Last Updated : 01 Jul, 2021

Given a Binary Tree and a key to be searched in it, write an iterative method that returns true if key is present in Binary Tree, else false.
For example, in the following tree, if the searched key is 3, then function should return true and if the searched key is 12, then function should return false.
 

1T

 

One thing is sure that we need to traverse complete tree to decide whether key is present or not. We can use any of the following traversals to iteratively search a key in a given binary tree. 
1) Iterative Level Order Traversal
2) Iterative Inorder Traversal 
3) Iterative Preorder Traversal 
4) Iterative Postorder Traversal
Below is iterative Level Order Traversal based solution to search an item x in binary tree. 
 

C++




// Iterative level order traversal
// based method to search in Binary Tree
#include<bits/stdc++.h>
 
using namespace std;
 
/* A binary tree node has data,
left child and right child */
class node
{
    public:
    int data;
    node* left;
    node* right;
     
    /* Constructor that allocates a new node with the
    given data and NULL left and right pointers. */
    node(int data){
        this->data = data;
        this->left = NULL;
        this->right = NULL;
         
    }
};
 
 
// An iterative process to search
// an element x in a given binary tree
bool iterativeSearch(node *root, int x)
{
    // Base Case
    if (root == NULL)
        return false;
 
    // Create an empty queue for
    // level order traversal
    queue<node *> q;
 
    // Enqueue Root and initialize height
    q.push(root);
 
    // Queue based level order traversal
    while (q.empty() == false)
    {
        // See if current node is same as x
        node *node = q.front();
        if (node->data == x)
            return true;
 
        // Remove current node and enqueue its children
        q.pop();
        if (node->left != NULL)
            q.push(node->left);
        if (node->right != NULL)
            q.push(node->right);
    }
 
    return false;
}
 
// Driver code
int main()
{
    node* NewRoot=NULL;
    node *root = new node(2);
    root->left = new node(7);
    root->right = new node(5);
    root->left->right = new node(6);
    root->left->right->left=new node(1);
    root->left->right->right=new node(11);
    root->right->right=new node(9);
    root->right->right->left=new node(4);
 
    iterativeSearch(root, 6)? cout <<
    "Found\n": cout << "Not Found\n";
    iterativeSearch(root, 12)? cout <<
    "Found\n": cout << "Not Found\n";
    return 0;
}
 
// This code is contributed by rathbhupendra

C




// Iterative level order traversal based method to search in Binary Tree
#include <iostream>
#include <queue>
using namespace std;
 
/* A binary tree node has data, left child and right child */
struct node
{
    int data;
    struct node* left, *right;
};
 
/* Helper function that allocates a new node with the given data and
   NULL left and right  pointers.*/
struct node* newNode(int data)
{
    struct node* node = new struct node;
    node->data = data;
    node->left = node->right = NULL;
    return(node);
}
 
// An iterative process to search an element x in a given binary tree
bool iterativeSearch(node *root, int x)
{
    // Base Case
    if (root == NULL)
        return false;
 
    // Create an empty queue for level order traversal
    queue<node *> q;
 
    // Enqueue Root and initialize height
    q.push(root);
 
    // Queue based level order traversal
    while (q.empty() == false)
    {
        // See if current node is same as x
        node *node = q.front();
        if (node->data == x)
            return true;
 
        // Remove current node and enqueue its children
        q.pop();
        if (node->left != NULL)
            q.push(node->left);
        if (node->right != NULL)
            q.push(node->right);
    }
 
    return false;
}
 
// Driver program
int main(void)
{
    struct node*NewRoot=NULL;
    struct node *root = newNode(2);
    root->left        = newNode(7);
    root->right       = newNode(5);
    root->left->right = newNode(6);
    root->left->right->left=newNode(1);
    root->left->right->right=newNode(11);
    root->right->right=newNode(9);
    root->right->right->left=newNode(4);
 
    iterativeSearch(root, 6)? cout << "Found\n": cout << "Not Found\n";
    iterativeSearch(root, 12)? cout << "Found\n": cout << "Not Found\n";
    return 0;
}

Java




// Iterative level order traversal
// based method to search in Binary Tree
import java.util.*;
 
class GFG
{
     
/* A binary tree node has data,
left child and right child */
static class node
{
    int data;
    node left;
    node right;
     
    /* Constructor that allocates a new node with the
    given data and null left and right pointers. */
    node(int data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
         
    }
};
 
 
// An iterative process to search
// an element x in a given binary tree
static boolean iterativeSearch(node root, int x)
{
    // Base Case
    if (root == null)
        return false;
 
    // Create an empty queue for
    // level order traversal
    Queue<node > q = new LinkedList();
 
    // Enqueue Root and initialize height
    q.add(root);
 
    // Queue based level order traversal
    while (q.size() > 0)
    {
        // See if current node is same as x
        node node = q.peek();
        if (node.data == x)
            return true;
 
        // Remove current node and enqueue its children
        q.remove();
        if (node.left != null)
            q.add(node.left);
        if (node.right != null)
            q.add(node.right);
    }
 
    return false;
}
 
// Driver code
public static void main(String ags[])
{
    node NewRoot = null;
    node root = new node(2);
    root.left = new node(7);
    root.right = new node(5);
    root.left.right = new node(6);
    root.left.right.left = new node(1);
    root.left.right.right = new node(11);
    root.right.right = new node(9);
    root.right.right.left = new node(4);
 
    System.out.print((iterativeSearch(root, 6)?
    "Found\n": "Not Found\n"));
    System.out.print((iterativeSearch(root, 12)?
    "Found\n": "Not Found\n"));
}
}
 
// This code is contributed by Arnab Kundu

Python3




# Iterative level order traversal based
# method to search in Binary Tree
 
# importing Queue
from queue import Queue
 
# Helper function that allocates a
# new node with the given data and
# None left and right pointers.
class newNode:
    def __init__(self, data):
        self.data = data
        self.left = self.right = None
 
# An iterative process to search an
# element x in a given binary tree
def iterativeSearch(root, x):
     
    # Base Case
    if (root == None):
        return False
 
    # Create an empty queue for level
    # order traversal
    q = Queue()
 
    # Enqueue Root and initialize height
    q.put(root)
 
    # Queue based level order traversal
    while (q.empty() == False):
         
        # See if current node is same as x
        node = q.queue[0]
        if (node.data == x):
            return True
 
        # Remove current node and
        # enqueue its children
        q.get()
        if (node.left != None):
            q.put(node.left)
        if (node.right != None):
            q.put(node.right)
 
    return False
 
# Driver Code
if __name__ == '__main__':
 
    root = newNode(2)
    root.left = newNode(7)
    root.right = newNode(5)
    root.left.right = newNode(6)
    root.left.right.left = newNode(1)
    root.left.right.right = newNode(11)
    root.right.right = newNode(9)
    root.right.right.left = newNode(4)
 
    if iterativeSearch(root, 6):
        print("Found")
    else:
        print("Not Found")
    if iterativeSearch(root, 12):
        print("Found")
    else:
        print("Not Found")
 
# This code is contributed by PranchalK

C#




// Iterative level order traversal
// based method to search in Binary Tree
using System;
using System.Collections.Generic;
 
class GFG
{
     
/* A binary tree node has data,
left child and right child */
public class node
{
    public int data;
    public node left;
    public node right;
     
    /* Constructor that allocates a new node
    with the given data and null left and
    right pointers. */
    public node(int data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
};
 
// An iterative process to search
// an element x in a given binary tree
static Boolean iterativeSearch(node root,
                               int x)
{
    // Base Case
    if (root == null)
        return false;
 
    // Create an empty queue for
    // level order traversal
    Queue<node > q = new Queue<node>();
 
    // Enqueue Root and initialize height
    q.Enqueue(root);
 
    // Queue based level order traversal
    while (q.Count > 0)
    {
        // See if current node is same as x
        node node = q.Peek();
        if (node.data == x)
            return true;
 
        // Remove current node and
        // enqueue its children
        q.Dequeue();
        if (node.left != null)
            q.Enqueue(node.left);
        if (node.right != null)
            q.Enqueue(node.right);
    }
    return false;
}
 
// Driver code
public static void Main(String []ags)
{
    node root = new node(2);
    root.left = new node(7);
    root.right = new node(5);
    root.left.right = new node(6);
    root.left.right.left = new node(1);
    root.left.right.right = new node(11);
    root.right.right = new node(9);
    root.right.right.left = new node(4);
 
    Console.WriteLine((iterativeSearch(root, 6) ?
                                      "Found\n" :
                                   "Not Found"));
    Console.Write((iterativeSearch(root, 12) ?
                                   "Found\n" :
                              "Not Found\n"));
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
// Iterative level order traversal
// based method to search in Binary Tree
 
/* A binary tree node has data,
left child and right child */
class node
{
    constructor(data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
 
// An iterative process to search
// an element x in a given binary tree
function iterativeSearch(root,x)
{
    // Base Case
    if (root == null)
        return false;
   
    // Create an empty queue for
    // level order traversal
    let q = [];
   
    // Enqueue Root and initialize height
    q.push(root);
   
    // Queue based level order traversal
    while (q.length > 0)
    {
        // See if current node is same as x
        let node = q[0];
        if (node.data == x)
            return true;
   
        // Remove current node and enqueue its children
        q.shift();
        if (node.left != null)
            q.push(node.left);
        if (node.right != null)
            q.push(node.right);
    }
   
    return false;
}
 
// Driver code
let NewRoot = null;
let root = new node(2);
root.left = new node(7);
root.right = new node(5);
root.left.right = new node(6);
root.left.right.left = new node(1);
root.left.right.right = new node(11);
root.right.right = new node(9);
root.right.right.left = new node(4);
 
document.write((iterativeSearch(root, 6)?
                  "Found<br>": "Not Found<br>"));
document.write((iterativeSearch(root, 12)?
                  "Found<br>": "Not Found<br>"));
 
// This code is contributed by rag2127
</script>

Output:  

Found
Not Found

  
Below implementation uses Iterative Preorder Traversal to find x in Binary Tree 
 
 

C++




// An iterative method to search an item in Binary Tree
#include <iostream>
#include <stack>
using namespace std;
 
/* A binary tree node has data, left child and right child */
struct node
{
    int data;
    struct node* left, *right;
};
 
/* Helper function that allocates a new node with the given data and
NULL left and right pointers.*/
struct node* newNode(int data)
{
    struct node* node = new struct node;
    node->data = data;
    node->left = node->right = NULL;
    return(node);
}
 
// iterative process to search an element x in a given binary tree
bool iterativeSearch(node *root, int x)
{
    // Base Case
    if (root == NULL)
        return false;
 
    // Create an empty stack and push root to it
    stack<node *> nodeStack;
    nodeStack.push(root);
 
    // Do iterative preorder traversal to search x
    while (nodeStack.empty() == false)
    {
        // See the top item from stack and check if it is same as x
        struct node *node = nodeStack.top();
        if (node->data == x)
            return true;
        nodeStack.pop();
 
        // Push right and left children of the popped node to stack
        if (node->right)
            nodeStack.push(node->right);
        if (node->left)
            nodeStack.push(node->left);
    }
 
    return false;
}
 
// Driver program
int main(void)
{
    struct node*NewRoot=NULL;
    struct node *root = newNode(2);
    root->left     = newNode(7);
    root->right     = newNode(5);
    root->left->right = newNode(6);
    root->left->right->left=newNode(1);
    root->left->right->right=newNode(11);
    root->right->right=newNode(9);
    root->right->right->left=newNode(4);
 
    iterativeSearch(root, 6)? cout << "Found\n": cout << "Not Found\n";
    iterativeSearch(root, 12)? cout << "Found\n": cout << "Not Found\n";
    return 0;
}

Java




// An iterative method to search an item in Binary Tree
import java.util.*;
 
class GFG
{
 
/* A binary tree node has data,
left child and right child */
static class node
{
    int data;
    node left, right;
};
 
/* Helper function that allocates a
new node with the given data and
null left and right pointers.*/
static node newNode(int data)
{
    node node = new node();
    node.data = data;
    node.left = node.right = null;
    return(node);
}
 
// iterative process to search
// an element x in a given binary tree
static boolean iterativeSearch(node root, int x)
{
    // Base Case
    if (root == null)
        return false;
 
    // Create an empty stack and push root to it
    Stack<node> nodeStack = new Stack<node>();
    nodeStack.push(root);
 
    // Do iterative preorder traversal to search x
    while (nodeStack.empty() == false)
    {
        // See the top item from stack and
        // check if it is same as x
        node node = nodeStack.peek();
        if (node.data == x)
            return true;
        nodeStack.pop();
 
        // Push right and left children
        // of the popped node to stack
        if (node.right != null)
            nodeStack.push(node.right);
        if (node.left != null)
            nodeStack.push(node.left);
    }
    return false;
}
 
// Driver Code
public static void main(String[] args)
{
    node NewRoot = null;
    node root = newNode(2);
    root.left = newNode(7);
    root.right = newNode(5);
    root.left.right = newNode(6);
    root.left.right.left = newNode(1);
    root.left.right.right = newNode(11);
    root.right.right = newNode(9);
    root.right.right.left = newNode(4);
 
    if(iterativeSearch(root, 6))
        System.out.println("Found");
    else
        System.out.println("Not Found");
    if(iterativeSearch(root, 12))
        System.out.println("Found");
    else
        System.out.println("Not Found");
}
}
 
// This code is contributed by 29AjayKumar

Python3




# An iterative Python3 code to search
# an item in Binary Tree
 
''' A binary tree node has data,
left child and right child '''
class newNode:
 
    # Construct to create a newNode
    def __init__(self, key):
        self.data = key
        self.left = None
        self.right = None
 
# iterative process to search an element x
# in a given binary tree
def iterativeSearch(root,x):
     
    # Base Case
    if (root == None):
        return False
     
    # Create an empty stack and
    # append root to it
    nodeStack = []
    nodeStack.append(root)
     
    # Do iterative preorder traversal to search x
    while (len(nodeStack)):
         
        # See the top item from stack and
        # check if it is same as x
        node = nodeStack[0]
        if (node.data == x):
            return True
        nodeStack.pop(0)
         
        # append right and left children
        # of the popped node to stack
        if (node.right):
            nodeStack.append(node.right)
        if (node.left):
            nodeStack.append(node.left)
     
    return False
 
# Driver Code
root = newNode(2)
root.left = newNode(7)
root.right = newNode(5)
root.left.right = newNode(6)
root.left.right.left = newNode(1)
root.left.right.right = newNode(11)
root.right.right = newNode(9)
root.right.right.left = newNode(4)
 
if iterativeSearch(root, 6):
    print("Found")
else:
    print("Not Found")
     
if iterativeSearch(root, 12):
    print("Found")
else:
    print("Not Found")
 
# This code is contributed by SHUBHAMSINGH10

C#




// An iterative method to search an item in Binary Tree
using System;
using System.Collections.Generic;
 
class GFG
{
 
/* A binary tree node has data,
left child and right child */
class node
{
    public int data;
    public node left, right;
};
 
/* Helper function that allocates a
new node with the given data and
null left and right pointers.*/
static node newNode(int data)
{
    node node = new node();
    node.data = data;
    node.left = node.right = null;
    return(node);
}
 
// iterative process to search
// an element x in a given binary tree
static bool iterativeSearch(node root, int x)
{
    // Base Case
    if (root == null)
        return false;
 
    // Create an empty stack and.Push root to it
    Stack<node> nodeStack = new Stack<node>();
    nodeStack.Push(root);
 
    // Do iterative preorder traversal to search x
    while (nodeStack.Count != 0)
    {
        // See the top item from stack and
        // check if it is same as x
        node node = nodeStack.Peek();
        if (node.data == x)
            return true;
        nodeStack.Pop();
 
        // Push right and left children
        // of the.Popped node to stack
        if (node.right != null)
            nodeStack.Push(node.right);
        if (node.left != null)
            nodeStack.Push(node.left);
    }
    return false;
}
 
// Driver Code
public static void Main(String[] args)
{
    node root = newNode(2);
    root.left = newNode(7);
    root.right = newNode(5);
    root.left.right = newNode(6);
    root.left.right.left = newNode(1);
    root.left.right.right = newNode(11);
    root.right.right = newNode(9);
    root.right.right.left = newNode(4);
 
    if(iterativeSearch(root, 6))
        Console.WriteLine("Found");
    else
        Console.WriteLine("Not Found");
    if(iterativeSearch(root, 12))
        Console.WriteLine("Found");
    else
        Console.WriteLine("Not Found");
}
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
  
// An iterative method to search an item in Binary Tree
 
/* A binary tree node has data,
left child and right child */
class Node
{
  constructor()
  {
    this.data = 0;
    this.left = null;
    this.right = null;
  }
};
 
/* Helper function that allocates a
new node with the given data and
null left and right pointers.*/
function newNode(data)
{
    var node = new Node();
    node.data = data;
    node.left = node.right = null;
    return(node);
}
 
// iterative process to search
// an element x in a given binary tree
function iterativeSearch(root, x)
{
    // Base Case
    if (root == null)
        return false;
 
    // Create an empty stack and.push root to it
    var nodeStack = [];
    nodeStack.push(root);
 
    // Do iterative preorder traversal to search x
    while (nodeStack.length != 0)
    {
        // See the top item from stack and
        // check if it is same as x
        var node = nodeStack[nodeStack.length - 1];
        if (node.data == x)
            return true;
        nodeStack.pop();
 
        // push right and left children
        // of the.Popped node to stack
        if (node.right != null)
            nodeStack.push(node.right);
        if (node.left != null)
            nodeStack.push(node.left);
    }
    return false;
}
 
// Driver Code
var root = newNode(2);
root.left = newNode(7);
root.right = newNode(5);
root.left.right = newNode(6);
root.left.right.left = newNode(1);
root.left.right.right = newNode(11);
root.right.right = newNode(9);
root.right.right.left = newNode(4);
if(iterativeSearch(root, 6))
    document.write("Found<br>");
else
    document.write("Not Found<br>");
if(iterativeSearch(root, 12))
    document.write("Found<br>");
else
    document.write("Not Found<br>");
 
</script>

Output:  

Found
Not Found

Similarly, Iterative Inorder and Iterative Postorder traversals can be used.
 

https://www.youtube.com/watch?v=-jREIGMJ8Tg
 

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

 


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