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

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. filter_none

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    // 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;
    }

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    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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