# Count of subtrees in a Binary Tree having XOR value K

• Last Updated : 22 Jun, 2021

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```

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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)

1.
2. If the XOR of any subtree is K, increment the counter variable.
3. Print the value in counter as the required count

Below is the implementation of the above approach:

## C++

 `// C++ program to find the count of``// subtrees in a Binary Tree``// having XOR value K` `#include ``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;``}`

## Java

 `// 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`

## Python3

 `# Python3 program to find the count of``# subtrees in a Binary Tree``# having XOR value K`` ` `# A binary tree node``class` `Node:``    ` `    ``def` `__init__(``self``, data):``        ` `        ``self``.data ``=` `data``        ``self``.left ``=` `None``        ``self``.right ``=` `None``        ` `# A utility function to``# allocate a new node``def` `newNode(data):``    ` `    ``newNode ``=` `Node(data)``    ``return` `newNode` `def` `rec(root, res, k):` `    ``# Base Case:``    ``# If node is None, return 0``    ``if` `(root ``=``=` `None``):``        ``return` `[``0``, res];`` ` `    ``# Calculating the XOR``    ``# of the current subtree``    ``xr ``=` `root.data;``    ``tmp,res ``=` `rec(root.left, res, k);``    ``xr^``=``tmp``    ``tmp,res ``=` `rec(root.right, res, k);``    ``xr^``=``tmp`` ` `    ``# Increment res``    ``# if xr is equal to k``    ``if` `(xr ``=``=` `k):``        ``res ``+``=` `1``  ` `    ``# Return the XOR value``    ``# of the current subtree``    ``return` `xr, res;` `# Function to find the required count``def` `findCount(root, K):` `    ``# Initialize result variable 'res'``    ``res ``=` `0``;`` ` `    ``# Recursively traverse the tree``    ``# and compute the count``    ``tmp,res``=``rec(root, res, K);`` ` `    ``# return the count 'res'``    ``return` `res;` `# Driver program``if` `__name__``=``=``'__main__'``:`` ` `    ``'''``        ``2``       ``/ \``      ``1   9``     ``/ \``    ``10  5``    ``'''`` ` `    ``# Create the binary tree``    ``# by adding nodes to it``    ``root ``=` `newNode(``2``);``    ``root.left ``=` `newNode(``1``);``    ``root.right ``=` `newNode(``9``);``    ``root.left.left ``=` `newNode(``10``);``    ``root.left.right ``=` `newNode(``5``);`` ` `    ``K ``=` `5``;`` ` `    ``print``(findCount(root, K))` `# This code is contributed by rutvik_56`

## C#

 `// 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`

## Javascript

 ``
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).

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