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Count of subtrees in a Binary Tree having bitwise OR value K
  • Last Updated : 12 Jan, 2021

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:  

  • 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++




// 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;
}

Java




// Java program to find the count of
// subtrees in a Binary Tree
// having bitwise OR value K
import java.io.*;
class GFG
{
   
    // A binary tree node
    static 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);
        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 code
    public static void main (String[] args)
    {
        /*
         2
        / \
       1   1
      / \   \
    10   5   4
    */
         
        // Create the binary tree
        // by adding nodes to it
        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;
        System.out.println(findCount(root, K));
    }
}
 
// This code is contributed by avanitrachhadiya2155

Python3




# Python3 program to find the count of
# subtrees in a Binary Tree
# having bitwise OR 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):
     
    temp = Node(data)
    return temp
   
# Recursive Function to compute the count
def rec(root, res, k):
 
    # Base Case:
    # If node is NULL, return 0
    if (root == None):
        return [0, res];
   
    # Calculating the bitwise OR
    # of the current subtree
    orr = root.data;
    tmp, res = rec(root.left, res, k);
    orr |= tmp
    tmp, res = rec(root.right, res, k);
    orr |= tmp
   
    # Increment res
    # if xr is equal to k
    if (orr == k):
        res += 1
   
    # Return the bitwise OR value
    # of the current subtree
    return orr, 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   1
    / \   \
   10  5   4
    '''
   
    # Create the binary tree
    # by adding nodes to it
    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);
   
    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 bitwise OR value K
using System;
 
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);
    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 code
public static void Main(String []args)
{
     
    /*
         2
        / \
       1   1
      / \   \
    10   5   4
    */
 
    // Create the binary tree
    // by adding nodes to it
    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;
 
    Console.WriteLine(findCount(root, K));
}
}
 
// This code is contributed by mohit kumar
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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