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:

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

bitwise OR of its subtree = (bitwise OR of node’s left subtree) | (bitwise OR of node’s right subtree) | (node’s value)

If the bitwise OR of any subtree is K, increment the counter variable.
Print the value in the counter as the required count.

// C++ program to find the count of 
// subtrees in a Binary Tree 
// having bitwise OR 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); 
} 
  
// 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:

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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My Personal Notes

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

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