Count of subtrees in a Binary Tree having XOR value K

Given a value K and a binary tree, the task is to find out number of subtrees having XOR of all its elements equal to K.

Examples:

Input  K = 5, Tree =
       2
      / \
     1   9
    / \
   10  5
Output: 2
Explanation:
Subtree 1: 
       5
It has only one element i.e. 5.
So XOR of subtree = 5
Subtree 1: 
       2
      / \
     1   9
    / \
   10  5
It has elements 2, 1, 9, 10, 5.
So XOR of subtree = 2 ^ 1 ^ 9 ^ 10 ^ 5 = 5

Input  K = 3, Tree =
       4
      / \
     3   9
    / \
   2   2
Output: 1
Explanation
Subtree:
     3
    / \
   2   2
It has elements 3, 2, 2.
So XOR of subtree = 3 ^ 2 ^ 2 = 3

Approach

  1. Traverse the tree recursively using pre-order traversal.
  2. For each node keep calculating the XOR of its subtree as:

    XOR of its subtree = (XOR of node’s left subtree) ^ (XOR of nodes’s right subtree) ^ (node’s value)

  3. If the XOR of any subtree is K, increment the counter variable.
  4. Print the value in counter as the required count

Below is the implementation of the above approach:

C++



filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find the count of
// subtrees in a Binary Tree
// having XOR value K
  
#include <bits/stdc++.h>
using namespace std;
  
// A binary tree node
struct Node {
    int data;
    struct Node *left, *right;
};
  
// A utility function to
// allocate a new node
struct Node* newNode(int data)
{
    struct Node* newNode = new Node;
    newNode->data = data;
    newNode->left
        = newNode->right = NULL;
    return (newNode);
}
  
int rec(Node* root, int& res, int& k)
{
    // Base Case:
    // If node is NULL, return 0
    if (root == NULL) {
        return 0;
    }
  
    // Calculating the XOR
    // of the current subtree
    int xr = root->data;
    xr ^= rec(root->left, res, k);
    xr ^= rec(root->right, res, k);
  
    // Increment res
    // if xr is equal to k
    if (xr == k) {
        res++;
    }
  
    // Return the XOR value
    // of the current subtree
    return xr;
}
  
// Function to find the required count
int findCount(Node* root, int K)
{
    // Initialize result variable 'res'
    int res = 0;
  
    // Recursively traverse the tree
    // and compute the count
    rec(root, res, K);
  
    // return the count 'res'
    return res;
}
  
// Driver program
int main(void)
{
  
    /* 
        2
       / \
      1   9
     / \
    10  5
    */
  
    // Create the binary tree
    // by adding nodes to it
    struct Node* root = newNode(2);
    root->left = newNode(1);
    root->right = newNode(9);
    root->left->left = newNode(10);
    root->left->right = newNode(5);
  
    int K = 5;
  
    cout << findCount(root, K);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to find the count of
// subtrees in a Binary Tree
// having XOR value K
import java.util.*;
  
class GFG{
  
    // A binary tree node
    static class Node
    
        int data; 
        Node left,right; 
    }; 
    static int res;
    static int k;
  
    // A utility function to
    // allocate a new node
    static Node newNode(int data)
    {
        Node newNode = new Node();
        newNode.data = data;
        newNode.left= null;
        newNode.right = null;
        return newNode;
    }
      
    static int rec(Node root)
    {
        // Base Case:
        // If node is null, return 0
        if (root == null) {
            return 0;
        }
      
        // Calculating the XOR
        // of the current subtree
        int xr = (root.data);//^rec(root.left)^rec(root.right);
        xr ^= rec(root.left);
        xr ^= rec(root.right);
          
        // Increment res
        // if xr is equal to k
        if (xr == k) {
            res++;
        }
      
        // Return the XOR value
        // of the current subtree
        return xr;
    }
      
    // Function to find the required count
    static int findCount(Node root, int K)
    {
        // Initialize result variable 'res'
        res = 0;
        k = K;
  
        // Recursively traverse the tree
        // and compute the count
        rec(root);
  
        // return the count 'res'
        return res;
    }
      
    // Driver program
    public static void main(String args[])
    {
      
        /* 
            2
        / \
        1 9
        / \
        10 5
        */
      
        // Create the binary tree
        // by adding nodes to it
        Node root = newNode(2);
        root.left = newNode(1);
        root.right = newNode(9);
        root.left.left =newNode(10);
        root.left.right = newNode(5);
      
        int K = 5;
      
        System.out.println(findCount(root, K));    
    }
}
  
  
// This code is contributed by AbhiThakur

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to find the count of
// subtrees in a Binary Tree
// having XOR value K
using System;
  
public class GFG{
   
    // A binary tree node
    class Node
    
        public int data; 
        public Node left,right; 
    }; 
    static int res;
    static int k;
   
    // A utility function to
    // allocate a new node
    static Node newNode(int data)
    {
        Node newNode = new Node();
        newNode.data = data;
        newNode.left= null;
        newNode.right = null;
        return newNode;
    }
       
    static int rec(Node root)
    {
        // Base Case:
        // If node is null, return 0
        if (root == null) {
            return 0;
        }
       
        // Calculating the XOR
        // of the current subtree
        int xr = (root.data);//^rec(root.left)^rec(root.right);
        xr ^= rec(root.left);
        xr ^= rec(root.right);
           
        // Increment res
        // if xr is equal to k
        if (xr == k) {
            res++;
        }
       
        // Return the XOR value
        // of the current subtree
        return xr;
    }
       
    // Function to find the required count
    static int findCount(Node root, int K)
    {
        // Initialize result variable 'res'
        res = 0;
        k = K;
   
        // Recursively traverse the tree
        // and compute the count
        rec(root);
   
        // return the count 'res'
        return res;
    }
       
    // Driver program
    public static void Main(String []args)
    {
       
        /* 
            2
        / \
        1 9
        / \
        10 5
        */
       
        // Create the binary tree
        // by adding nodes to it
        Node root = newNode(2);
        root.left = newNode(1);
        root.right = newNode(9);
        root.left.left =newNode(10);
        root.left.right = newNode(5);
       
        int K = 5;
       
        Console.WriteLine(findCount(root, K));    
    }
}
   
// This code is contributed by sapnasingh4991

chevron_right


Output:

2

Performance Analysis:

Time Complexity: As in the above approach, we are iterating over each node only once, therefore it will take O(N) time where N is the number of nodes in the Binary tree.

Auxiliary Space Complexity: As in the above approach there is no extra space used, therefore the Auxiliary Space complexity will be O(1).

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.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

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.