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Search a node in Binary Tree

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  • Difficulty Level : Easy
  • Last Updated : 13 Jun, 2022

Given a Binary tree and a node. The task is to search and check if the given node exists in the binary tree or not. If it exists, print YES otherwise print NO.
Given Binary Tree
 

Examples

Input: Node = 4
Output: YES

Input: Node = 40
Output: NO

The idea is to use any of the tree traversals to traverse the tree and while traversing check if the current node matches with the given node. Print YES if any node matches with the given node and stop traversing further and if the tree is completely traversed and none of the node matches with the given node then print NO.

Below is the implementation of the above approach: 

C++




// C++ program to check if a node exists
// in a binary tree
#include <iostream>
using namespace std;
 
// Binary tree node
struct Node {
    int data;
    struct Node *left, *right;
    Node(int data)
    {
        this->data = data;
        left = right = NULL;
    }
};
 
// Function to traverse the tree in preorder
// and check if the given node exists in it
bool ifNodeExists(struct Node* node, int key)
{
    if (node == NULL)
        return false;
 
    if (node->data == key)
        return true;
 
    /* then recur on left subtree */
    bool res1 = ifNodeExists(node->left, key);
    // node found, no need to look further
    if(res1) return true;
 
    /* node is not found in left,
    so recur on right subtree */
    bool res2 = ifNodeExists(node->right, key);
 
    return res2;
}
 
// Driver Code
int main()
{
    struct Node* root = new Node(0);
    root->left = new Node(1);
    root->left->left = new Node(3);
    root->left->left->left = new Node(7);
    root->left->right = new Node(4);
    root->left->right->left = new Node(8);
    root->left->right->right = new Node(9);
    root->right = new Node(2);
    root->right->left = new Node(5);
    root->right->right = new Node(6);
 
    int key = 4;
 
    if (ifNodeExists(root, key))
        cout << "YES";
    else
        cout << "NO";
 
    return 0;
}

Java




// Java program to check if a node exists
// in a binary tree
class GFG
{
     
// Binary tree node
static class Node
{
    int data;
    Node left, right;
    Node(int data)
    {
        this.data = data;
        left = right = null;
    }
};
 
// Function to traverse the tree in preorder
// and check if the given node exists in it
static boolean ifNodeExists( Node node, int key)
{
    if (node == null)
        return false;
 
    if (node.data == key)
        return true;
 
    // then recur on left subtree /
    boolean res1 = ifNodeExists(node.left, key);
   
    // node found, no need to look further
    if(res1) return true;
 
    // node is not found in left,
    // so recur on right subtree /
    boolean res2 = ifNodeExists(node.right, key);
 
    return res2;
}
 
// Driver Code
public static void main(String args[])
{
    Node root = new Node(0);
    root.left = new Node(1);
    root.left.left = new Node(3);
    root.left.left.left = new Node(7);
    root.left.right = new Node(4);
    root.left.right.left = new Node(8);
    root.left.right.right = new Node(9);
    root.right = new Node(2);
    root.right.left = new Node(5);
    root.right.right = new Node(6);
 
    int key = 4;
 
    if (ifNodeExists(root, key))
        System.out.println("YES");
    else
        System.out.println("NO");
}
}
 
// This code is contributed by Arnab Kundu

Python3




"""Python program to check if a node exists
in a binary tree."""
 
# A Binary Tree Node
# Utility function to create a new tree node
class newNode:
 
    # Constructor to create a newNode
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
 
# Function to traverse the tree in preorder
# and check if the given node exists in it
def ifNodeExists(node, key):
 
    if (node == None):
        return False
 
    if (node.data == key):
        return True
 
    """ then recur on left subtree """
    res1 = ifNodeExists(node.left, key)
    # node found, no need to look further
    if res1:
        return True
 
    """ node is not found in left,
    so recur on right subtree """
    res2 = ifNodeExists(node.right, key)
 
    return res2
     
# Driver Code
if __name__ == '__main__':
    root = newNode(0)
    root.left = newNode(1)
    root.left.left = newNode(3)
    root.left.left.left = newNode(7)
    root.left.right = newNode(4)
    root.left.right.left = newNode(8)
    root.left.right.right = newNode(9)
    root.right = newNode(2)
    root.right.left = newNode(5)
    root.right.right = newNode(6)
 
    key = 4
 
    if (ifNodeExists(root, key)):
        print("YES" )
    else:
        print("NO")
 
# This code is contributed by SHUBHAMSINGH10

C#




// C# program to check if a node exists
// in a binary tree
using System;
     
class GFG
{
      
// Binary tree node
public class Node
{
    public int data;
    public Node left, right;
    public Node(int data)
    {
        this.data = data;
        left = right = null;
    }
};
  
// Function to traverse the tree in preorder
// and check if the given node exists in it
static bool ifNodeExists( Node node, int key)
{
    if (node == null)
        return false;
  
    if (node.data == key)
        return true;
  
    // then recur on left subtree /
    bool res1 = ifNodeExists(node.left, key);
   
    // node found, no need to look further
    if(res1) return true;
 
    // node is not found in left,
    // so recur on right subtree /
    bool res2 = ifNodeExists(node.right, key);
  
    return res2;
}
  
// Driver Code
public static void Main(String []args)
{
    Node root = new Node(0);
    root.left = new Node(1);
    root.left.left = new Node(3);
    root.left.left.left = new Node(7);
    root.left.right = new Node(4);
    root.left.right.left = new Node(8);
    root.left.right.right = new Node(9);
    root.right = new Node(2);
    root.right.left = new Node(5);
    root.right.right = new Node(6);
  
    int key = 4;
  
    if (ifNodeExists(root, key))
        Console.WriteLine("YES");
    else
       Console.WriteLine("NO");
}
}
 
// This code has been contributed by 29AjayKumar

Javascript




<script>
// javascript program to check if a node exists
// in a binary tree     // Binary tree node
     class Node {
         
         constructor(data) {
            this.data = data;
            this.left = this.right = null;
        }
    }
 
    // Function to traverse the tree in preorder
    // and check if the given node exists in it
    function ifNodeExists(node , key) {
        if (node == null)
            return false;
 
        if (node.data == key)
            return true;
 
        // then recur on left subtree /
        var res1 = ifNodeExists(node.left, key);
 
        // node found, no need to look further
        if (res1)
            return true;
 
        // node is not found in left,
        // so recur on right subtree /
        var res2 = ifNodeExists(node.right, key);
 
        return res2;
    }
 
    // Driver Code
     
var root = new Node(0);
        root.left = new Node(1);
        root.left.left = new Node(3);
        root.left.left.left = new Node(7);
        root.left.right = new Node(4);
        root.left.right.left = new Node(8);
        root.left.right.right = new Node(9);
        root.right = new Node(2);
        root.right.left = new Node(5);
        root.right.right = new Node(6);
 
        var key = 4;
 
        if (ifNodeExists(root, key))
            document.write("YES");
        else
            document.write("NO");
 
// This code is contributed by todaysgaurav
</script>

Output

YES

Time Complexity: O(N), as we are using recursion to traverse N nodes of the tree.

Auxiliary Space: O(N), we are not using any explicit extra space but as we are using recursion there will be extra space allocated for recursive stack.


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