Search a node in Binary Tree

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: 

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 sutree */
    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;
}
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 sutree /
    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
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

"""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 sutree """
    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
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 sutree /
    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
chevron_right

Output
YES

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.





Striver(underscore)79 at Codechef and codeforces D

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Article Tags :
Practice Tags :