Skip to content
Related Articles

Related Articles

Find the level with maximum setbit count in given Binary Tree

View Discussion
Improve Article
Save Article
  • Difficulty Level : Medium
  • Last Updated : 05 Jul, 2022

Given a binary tree having N nodes, the task is to find the level having the maximum number of setbits.

Note: If two levels have same number of setbits print the one which has less no of nodes. If nodes are equal print the first level from top to bottom

Examples: 

Input:      
         2
      /   \
    5     3
  /  \
6   1
Output: 2
Explanation: Level 1 has only one setbit  => 2 (010).
Level 2 has 4 setbits. => 5 (101) + 3 (011). 
Level 3 has 3 setbits. => 6 (110) +1 (001).

Input: 
          2
       /    \
     5      3
  /  \       \
6   1        8
Output: 2

 

Approach: The problem can be solved using level order traversal itself. Find the number of setbits in each level and the level having the maximum number of setbits following the given condition in the problem. Follow the steps mentioned below:

  • Use the level order traversal and for each level:
    • Find the total number of setbits in each level.
    • Update the maximum setbits in a level and the level having the maximum number of setbits.
  • Return the level with maximum setbits.

Below is the implementation of the above approach.

C++




// C++ code to implement the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Structure of a binary tree node
struct Node {
    int data;
    struct Node *left, *right;
};
 
// Function to count no of set bit
int countSetBit(int x)
{
    int c = 0;
 
    while (x) {
        int l = x % 10;
        if (x & 1)
            c++;
        x /= 2;
    }
    return c;
}
 
// Function to convert tree element
// by count of set bit they have
void convert(Node* root)
{
    if (!root)
        return;
    root->data = countSetBit(root->data);
    convert(root->left);
    convert(root->right);
}
 
// Function to get level with max set bit
int printLevel(Node* root)
{
    // Base Case
    if (root == NULL)
        return 0;
 
    // Replace tree elements by
    // count of set bits they contain
    convert(root);
 
    // Create an empty queue
    queue<Node*> q;
 
    int currLevel = 0, ma = INT_MIN;
    int prev = 0, ans = 0;
 
    // Enqueue Root and initialize height
    q.push(root);
 
    // Loop to implement level order traversal
    while (q.empty() == false) {
 
        // Print front of queue and
        // remove it from queue
        int size = q.size();
        currLevel++;
        int totalSet = 0, nodeCount = 0;
 
        while (size--) {
            Node* node = q.front();
 
            // Add all the set bit
            // in the current level
            totalSet += node->data;
            q.pop();
 
            // Enqueue left child
            if (node->left != NULL)
                q.push(node->left);
 
            // Enqueue right child
            if (node->right != NULL)
                q.push(node->right);
 
            // Count current level node
            nodeCount++;
        }
 
        // Update the ans when needed
        if (ma < totalSet) {
            ma = totalSet;
            ans = currLevel;
        }
 
        // If two level have same set bit
        // one with less node become ans
        else if (ma == totalSet && prev > nodeCount) {
            ma = totalSet;
            ans = currLevel;
            prev = nodeCount;
        }
 
        // Assign prev =
        // current level node count
        // We can use it for further levels
        // When 2 level have
        // same set bit count
        // print level with less node
        prev = nodeCount;
    }
    return ans;
}
 
// Utility function to create new tree node
Node* newNode(int data)
{
    Node* temp = new Node;
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}
 
// Driver program
int main()
{
    // Binary tree as shown in example
    Node* root = newNode(2);
    root->left = newNode(5);
    root->right = newNode(3);
    root->left->left = newNode(6);
    root->left->right = newNode(1);
    root->right->right = newNode(8);
 
    // Function call
    cout << printLevel(root) << endl;
    return 0;
}


Java




// Java code to implement the above approach
import java.util.*;
 
class GFG{
 
  // Structure of a binary tree node
  static class Node {
    int data;
    Node left, right;
  };
 
  // Function to count no of set bit
  static int countSetBit(int x)
  {
    int c = 0;
 
    while (x!=0) {
      int l = x % 10;
      if (x%2==1)
        c++;
      x /= 2;
    }
    return c;
  }
 
  // Function to convert tree element
  // by count of set bit they have
  static void convert(Node root)
  {
    if (root==null)
      return;
    root.data = countSetBit(root.data);
    convert(root.left);
    convert(root.right);
  }
 
  // Function to get level with max set bit
  static int printLevel(Node root)
  {
    // Base Case
    if (root == null)
      return 0;
 
    // Replace tree elements by
    // count of set bits they contain
    convert(root);
 
    // Create an empty queue
    Queue<Node> q = new LinkedList<>();
 
    int currLevel = 0, ma = Integer.MIN_VALUE;
    int prev = 0, ans = 0;
 
    // Enqueue Root and initialize height
    q.add(root);
 
    // Loop to implement level order traversal
    while (q.isEmpty() == false) {
 
      // Print front of queue and
      // remove it from queue
      int size = q.size();
      currLevel++;
      int totalSet = 0, nodeCount = 0;
 
      while (size-- >0) {
        Node node = q.peek();
 
        // Add all the set bit
        // in the current level
        totalSet += node.data;
        q.remove();
 
        // Enqueue left child
        if (node.left != null)
          q.add(node.left);
 
        // Enqueue right child
        if (node.right != null)
          q.add(node.right);
 
        // Count current level node
        nodeCount++;
      }
 
      // Update the ans when needed
      if (ma < totalSet) {
        ma = totalSet;
        ans = currLevel;
      }
 
      // If two level have same set bit
      // one with less node become ans
      else if (ma == totalSet && prev > nodeCount) {
        ma = totalSet;
        ans = currLevel;
        prev = nodeCount;
      }
 
      // Assign prev =
      // current level node count
      // We can use it for further levels
      // When 2 level have
      // same set bit count
      // print level with less node
      prev = nodeCount;
    }
    return ans;
  }
 
  // Utility function to create new tree node
  static Node newNode(int data)
  {
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
  }
 
  // Driver program
  public static void main(String[] args)
  {
    // Binary tree as shown in example
    Node root = newNode(2);
    root.left = newNode(5);
    root.right = newNode(3);
    root.left.left = newNode(6);
    root.left.right = newNode(1);
    root.right.right = newNode(8);
 
    // Function call
    System.out.print(printLevel(root) +"\n");
  }
}
 
// This code is contributed by shikhasingrajput


Python3




# Python code for the above approach
 
# Structure of a binary Tree node
import sys
class Node:
    def __init__(self,d):
        self.data = d
        self.left = None
        self.right = None
 
# Function to count no of set bit
def countSetBit(x):
    c = 0
 
    while (x):
        l = x % 10
        if (x & 1):
            c += 1
        x = (x // 2)
    return c
 
 # Function to convert tree element
 # by count of set bit they have
def convert(root):
    if (root == None):
        return
    root.data = countSetBit(root.data)
    convert(root.left)
    convert(root.right)
 
 # Function to get level with max set bit
def printLevel(root):
    # Base Case
    if (root == None):
        return 0
 
    # Replace tree elements by
    # count of set bits they contain
    convert(root)
 
    # Create an empty queue
    q = []
 
    currLevel,ma = 0, -sys.maxsize - 1
    prev,ans = 0,0
 
    # Enqueue Root and initialize height
    q.append(root)
 
    # Loop to implement level order traversal
    while (len(q) != 0):
 
        # Print front of queue and
        # remove it from queue
        size = len(q)
        currLevel += 1
        totalSet,nodeCount = 0,0
 
        while (size):
            node = q[0]
            q = q[1:]
 
            # Add all the set bit
            # in the current level
            totalSet += node.data
 
            # Enqueue left child
            if (node.left != None):
                q.append(node.left)
 
            # Enqueue right child
            if (node.right != None):
                q.append(node.right)
 
            # Count current level node
            nodeCount += 1
            size -= 1
 
        # Update the ans when needed
        if (ma < totalSet):
            ma = totalSet
            ans = currLevel
 
        # If two level have same set bit
        # one with less node become ans
        elif (ma == totalSet and prev > nodeCount):
            ma = totalSet
            ans = currLevel
            prev = nodeCount
   
        # Assign prev =
        # current level node count
        # We can use it for further levels
        # When 2 level have
        # same set bit count
        # print level with less node
        prev = nodeCount
    return ans
 
# Driver program
 
# Binary tree as shown in example
root = Node(2)
root.left = Node(5)
root.right = Node(3)
root.left.left = Node(6)
root.left.right = Node(1)
root.right.right = Node(8)
 
# Function call
print(printLevel(root))
 
# This code is contributed by shinjanpatra


C#




// C# code to implement the above approach
using System;
using System.Collections.Generic;
 
public class GFG{
 
  // Structure of a binary tree node
  class Node {
    public int data;
    public Node left, right;
  };
 
  // Function to count no of set bit
  static int countSetBit(int x)
  {
    int c = 0;
 
    while (x!=0) {
      int l = x % 10;
      if (x%2==1)
        c++;
      x /= 2;
    }
    return c;
  }
 
  // Function to convert tree element
  // by count of set bit they have
  static void convert(Node root)
  {
    if (root==null)
      return;
    root.data = countSetBit(root.data);
    convert(root.left);
    convert(root.right);
  }
 
  // Function to get level with max set bit
  static int printLevel(Node root)
  {
    // Base Case
    if (root == null)
      return 0;
 
    // Replace tree elements by
    // count of set bits they contain
    convert(root);
 
    // Create an empty queue
    Queue<Node> q = new Queue<Node>();
 
    int currLevel = 0, ma = int.MinValue;
    int prev = 0, ans = 0;
 
    // Enqueue Root and initialize height
    q.Enqueue(root);
 
    // Loop to implement level order traversal
    while (q.Count!=0 ) {
 
      // Print front of queue and
      // remove it from queue
      int size = q.Count;
      currLevel++;
      int totalSet = 0, nodeCount = 0;
 
      while (size-- >0) {
        Node node = q.Peek();
 
        // Add all the set bit
        // in the current level
        totalSet += node.data;
        q.Dequeue();
 
        // Enqueue left child
        if (node.left != null)
          q.Enqueue(node.left);
 
        // Enqueue right child
        if (node.right != null)
          q.Enqueue(node.right);
 
        // Count current level node
        nodeCount++;
      }
 
      // Update the ans when needed
      if (ma < totalSet) {
        ma = totalSet;
        ans = currLevel;
      }
 
      // If two level have same set bit
      // one with less node become ans
      else if (ma == totalSet && prev > nodeCount) {
        ma = totalSet;
        ans = currLevel;
        prev = nodeCount;
      }
 
      // Assign prev =
      // current level node count
      // We can use it for further levels
      // When 2 level have
      // same set bit count
      // print level with less node
      prev = nodeCount;
    }
    return ans;
  }
 
  // Utility function to create new tree node
  static Node newNode(int data)
  {
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
  }
 
  // Driver program
  public static void Main(String[] args)
  {
    // Binary tree as shown in example
    Node root = newNode(2);
    root.left = newNode(5);
    root.right = newNode(3);
    root.left.left = newNode(6);
    root.left.right = newNode(1);
    root.right.right = newNode(8);
 
    // Function call
    Console.Write(printLevel(root) +"\n");
  }
}
 
 
 
// This code contributed by shikhasingrajput


Javascript




<script>
        // JavaScript code for the above approach
 
 
        // Structure of a binary Tree node
        class Node {
            constructor(d) {
                this.data = d;
                this.left = null;
                this.right = null;
            }
        };
 
        // Function to count no of set bit
        function countSetBit(x) {
            let c = 0;
 
            while (x) {
                let l = x % 10;
                if (x & 1)
                    c++;
                x = Math.floor(x / 2);
            }
            return c;
        }
 
        // Function to convert tree element
        // by count of set bit they have
        function convert(root) {
            if (root == null)
                return;
            root.data = countSetBit(root.data);
            convert(root.left);
            convert(root.right);
        }
 
        // Function to get level with max set bit
        function printLevel(root) {
            // Base Case
            if (root == null)
                return 0;
 
            // Replace tree elements by
            // count of set bits they contain
            convert(root);
 
            // Create an empty queue
            let q = [];
 
            let currLevel = 0, ma = Number.MIN_VALUE;
            let prev = 0, ans = 0;
 
            // Enqueue Root and initialize height
            q.push(root);
 
            // Loop to implement level order traversal
            while (q.length != 0) {
 
                // Print front of queue and
                // remove it from queue
                let size = q.length;
                currLevel++;
                let totalSet = 0, nodeCount = 0;
 
                while (size--) {
                    let node = q.shift();
 
                    // Add all the set bit
                    // in the current level
                    totalSet += node.data;
                    q.pop();
 
                    // Enqueue left child
                    if (node.left != null)
                        q.push(node.left);
 
                    // Enqueue right child
                    if (node.right != null)
                        q.push(node.right);
 
                    // Count current level node
                    nodeCount++;
                }
 
                // Update the ans when needed
                if (ma < totalSet) {
                    ma = totalSet;
                    ans = currLevel;
                }
 
                // If two level have same set bit
                // one with less node become ans
                else if (ma == totalSet && prev > nodeCount) {
                    ma = totalSet;
                    ans = currLevel;
                    prev = nodeCount;
                }
 
                // Assign prev =
                // current level node count
                // We can use it for further levels
                // When 2 level have
                // same set bit count
                // print level with less node
                prev = nodeCount;
            }
            return ans;
        }
 
        // Driver program
 
        // Binary tree as shown in example
        let root = new Node(2);
        root.left = new Node(5);
        root.right = new Node(3);
        root.left.left = new Node(6);
        root.left.right = new Node(1);
        root.right.right = new Node(8);
 
        // Function call
        document.write(printLevel(root) + '<br>');
 
       // This code is contributed by Potta Lokesh
    </script>


Output

2

Time Complexity: (N * D) where D is no of bit an element have
Auxiliary Space: O(N)


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!