Find the level with maximum setbit count in given Binary Tree
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 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
# 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# 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
<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>
Time Complexity: (N * D) where D is no of bit an element have
Auxiliary Space: O(N)
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