Print the number of set bits in each node of a Binary Tree in C++ Programming.

C++Server Side ProgrammingProgramming

Given the binary tree, the function will generate the binary values of the keys stored in the nodes and then return the number of set bits(1) in that binary equivalent.

Example

Binary tree having keys as: 10 3 211 140 162 100 and 146

KeyBinary equivalentSet bits(output)
1010102
300112
211110100115
140100011003
162101000103
10011001003
146100100103

Here we are using the function __builtin_popcount

The function prototype is as follows −

int __builtin_popcount(unsigned int)

It returns the numbers of set bits in an integer i.e. the number of ones in the binary representation of the integer.

Algorithm

START
Step 1 -> create a structure of a node as
   struct Node
      struct node *left, *right
      int data
   End
Step 2 -> function to create a node
   node* newnode(int data)
   node->data = data
   node->left = node->right = NULL;
   return (node)
Step 3 -> Create function for generating bits of a node data
   void bits(Node* root)
   IF root = NULL
      return
   print __builtin_popcount(root->data)
   bits(root->left)
   bits(root->right)
step 4 -> In main()
   create tree using Node* root = newnode(10)
   root->left = newnode(3)
   call bits(root)
STOP

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
// structure of a node
struct Node {
   int data;
   struct Node *left, *right;
};
//function to create a new node
Node* newnode(int data) {
   Node* node = new Node;
   node->data = data;
   node->left = node->right = NULL;
   return (node);
}
//function for finding out the node
void bits(Node* root){
   if (root == NULL)
   return;
   //__builtin_popcount counts the number of set bit of a current node
   cout << "bits in node " << root->data << " = " <<__builtin_popcount(root->data)<< "\n";
   bits(root->left);
   bits(root->right);
}
int main(){
   Node* root = newnode(10);
   root->left = newnode(3);
   root->left->left = newnode(140);
   root->left->right = newnode(162);
   root->right = newnode(211);
   root->right->left = newnode(100);
   root->right->right = newnode(146);
   bits(root);
   return 0;
}

Output

if we run the above program then it will generate the following output

bits in node 10 = 2
bits in node 3 = 2
bits in node 140 = 3
bits in node 162 = 3
bits in node 211 = 5
bits in node 100 = 3
bits in node 146 = 3
raja
Published on 04-Sep-2019 06:29:29
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