Count of subtrees in a Binary Tree having bitwise OR value K

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

Examples:

```Input: K = 5, Tree = 2
/ \
1   1
/ \   \
10  5   4

Output:  2

Explanation:
Subtree 1:
5
It has only one element i.e. 5.
So bitwise OR of subtree = 5

Subtree 2:
1
\
4
it has 2 elements and bitwise OR of them is also 5

Input: K = 3, Tree =   4
/ \
3   9
/ \
2   2

Output:  1
```

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

1. Traverse the tree recursively using pre-order traversal.
2. For each node keep calculating the bitwise OR of its subtree as:
3. bitwise OR of its subtree = (bitwise OR of node’s left subtree) | (bitwise OR of node’s right subtree) | (node’s value)

4. If the bitwise OR of any subtree is K, increment the counter variable.
5. Print the value in the counter as the required count.
6.  `// C++ program to find the count of ` `// subtrees in a Binary Tree ` `// having bitwise OR 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); ` `} ` ` `  `// Recursive Function to compute the count ` `int` `rec(Node* root, ``int``& res, ``int``& k) ` `{ ` `    ``// Base Case: ` `    ``// If node is NULL, return 0 ` `    ``if` `(root == NULL) { ` `        ``return` `0; ` `    ``} ` ` `  `    ``// Calculating the bitwise OR ` `    ``// of the current subtree ` `    ``int` `orr = root->data; ` `    ``orr |= rec(root->left, res, k); ` `    ``orr |= rec(root->right, res, k); ` ` `  `    ``// Increment res ` `    ``// if xr is equal to k ` `    ``if` `(orr == k) { ` `        ``res++; ` `    ``} ` ` `  `    ``// Return the bitwise OR value ` `    ``// of the current subtree ` `    ``return` `orr; ` `} ` ` `  `// 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   1 ` `    ``/ \   \ ` `   ``10  5   4 ` `    ``*/` ` `  `    ``// Create the binary tree ` `    ``// by adding nodes to it ` `    ``struct` `Node* root = newNode(2); ` `    ``root->left = newNode(1); ` `    ``root->right = newNode(1); ` `    ``root->right->right = newNode(4); ` `    ``root->left->left = newNode(10); ` `    ``root->left->right = newNode(5); ` ` `  `    ``int` `K = 5; ` ` `  `    ``cout << FindCount(root, K); ` `    ``return` `0; ` `} `

Output:

```2
```

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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