Maximize count of set bits in a root to leaf path in a binary tree

Given a binary tree, the task is to find the total count of set bits in the node values of all the root to leaf paths and print the maximum among them.

Examples:

Input:

Output: 12
Explanation:
Path 1: 15(1111)->3(0011)->5(0101) = 8
Path 2: 15(1111)->3(0011)->1(0001) = 7
Path 3: 15(01111)->7(00111)->31(11111) = 12 (maximum)
Path 4: 15(1111)->7(0111)->9(1001) = 9
Therefore, the maximum count of set bits obtained in a path is 12.

Input:

Output: 13
Explanation:
Path 1: 31(11111)->3(00011)->7(00111) = 10
Path 2: 31(11111)->3(00011)->1(00001) = 8
Path 3: 31(11111)->15(01111)->5(00101) = 11
Path 4: 31(11111)->15(01111)->23(10111) = 13 (maximum)
Therefore, the maximum count of set bits obtained in a path is 13.

Approach:
Follow the steps below to solve the problem:

Traverse each node recursively, starting from the root node
Calculate the number of set bits in the value of the current node.
Update the maximum count of set bits(stored in a variable, say maxm).
Traverse its left and right subtree.
After complete traversal of all the nodes of the tree, print the final value of maxm as the answer.

Below is the implementation of the above approach:

C++

// C++ Program to implement 
// the above approach 
#include <bits/stdc++.h> 
using namespace std; 
int maxm = 0; 
 
// Node structure 
struct Node { 
    int val; 
 
    // Pointers to left 
    // and right child 
    Node *left, *right; 
 
    // Intialize consutructor 
    Node(int x) 
    { 
        val = x; 
        left = NULL; 
        right = NULL; 
    } 
}; 
 
// Function to find the maximum 
// count of setbits in a root to leaf 
void maxm_setbits(Node* root, int ans) 
{ 
    // Check if root is not null 
    if (!root) 
        return; 
 
    if (root->left == NULL 
        && root->right == NULL) { 
 
        ans += __builtin_popcount(root->val); 
 
        // Update the maximum count 
        // of setbits 
        maxm = max(ans, maxm); 
 
        return; 
    } 
 
    // Traverse left of binary tree 
    maxm_setbits(root->left, 
                ans + __builtin_popcount( 
                        root->val)); 
 
    // Traverse right of the binary tree 
    maxm_setbits(root->right, 
                ans + __builtin_popcount( 
                        root->val)); 
} 
 
// Driver Code 
int main() 
{ 
    Node* root = new Node(15); 
    root->left = new Node(3); 
    root->right = new Node(7); 
    root->left->left = new Node(5); 
    root->left->right = new Node(1); 
    root->right->left = new Node(31); 
    root->right->right = new Node(9); 
 
    maxm_setbits(root, 0); 
 
    cout << maxm << endl; 
 
    return 0; 
} 

Java

// Java Program to implement 
// the above approach 
import java.util.*;
class GFG{ 
   
static int maxm = 0; 
 
// Node structure 
static class Node 
{ 
    int val; 
 
    // Pointers to left 
    // and right child 
    Node left, right; 
 
    // Intialize consutructor 
    Node(int x) 
    { 
        val = x; 
        left = null; 
        right = null; 
    } 
}; 
 
// Function to find the maximum 
// count of setbits in a root to leaf 
static void maxm_setbits(Node root, int ans) 
{ 
    // Check if root is not null 
    if (root == null) 
        return; 
 
    if (root.left == null && 
        root.right == null) 
    { 
        ans += Integer.bitCount(root.val); 
 
        // Update the maximum count 
        // of setbits 
        maxm = Math.max(ans, maxm); 
 
        return; 
    } 
 
    // Traverse left of binary tree 
    maxm_setbits(root.left, 
                ans + Integer.bitCount( 
                        root.val)); 
 
    // Traverse right of the binary tree 
    maxm_setbits(root.right, 
                ans + Integer.bitCount( 
                        root.val)); 
} 
 
// Driver Code 
public static void main(String[] args) 
{ 
    Node root = new Node(15); 
    root.left = new Node(3); 
    root.right = new Node(7); 
    root.left.left = new Node(5); 
    root.left.right = new Node(1); 
    root.right.left = new Node(31); 
    root.right.right = new Node(9); 
 
    maxm_setbits(root, 0); 
 
    System.out.print(maxm +"\n"); 
 
} 
} 
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program to implement 
# the above approach
maxm = 0
 
# Node class
class Node:
     
    # Initialise constructor
    def __init__(self, x):
         
        self.val = x
        self.left = None
        self.right = None
         
# Function to count the number of 1 in number
def count_1(n):
     
    count = 0
    while (n): 
        count += n & 1
        n >>= 1
         
    return count 
 
# Function to find the maximum 
# count of setbits in a root to leaf
def maxm_setbits(root, ans):
     
    global maxm
     
    # Check if root is null
    if not root:
        return
     
    if (root.left == None and
        root.right == None):
        ans += count_1(root.val)
         
        # Update the maximum count 
        # of setbits 
        maxm = max(ans, maxm)
        return
     
    # Traverse left of binary tree
    maxm_setbits(root.left, 
                 ans + count_1(root.val))
     
    # Traverse right of the binary tree 
    maxm_setbits(root.right, 
                 ans + count_1(root.val))
     
# Driver code
root = Node(15)
root.left = Node(3)
root.right = Node(7)
root.left.left = Node(5)
root.left.right = Node(1)
root.right.left = Node(31)
root.right.right = Node(9)
 
maxm_setbits(root, 0)
 
print(maxm)
         
# This code is contributed by Stuti Pathak

C#

// C# program for the above approach 
using System;
class GFG{
     
// Function to Sort a Bitonic array 
// in constant space 
static void sortArr(int []a, int n) 
{ 
    int i, k; 
 
    // Initialse the value of k 
    k = (int)(Math.Log(n) / Math.Log(2)); 
    k = (int) Math.Pow(2, k); 
 
    // In each iteration compare elements 
    // k distance apart and swap if 
    // they are not in order 
    while (k > 0) 
    { 
        for(i = 0; i + k < n; i++) 
            if (a[i] > a[i + k])
            {
                int tmp = a[i];
                a[i] = a[i + k];
                a[i + k] = tmp;
            }
 
        // k is reduced to half 
        // after every iteration 
        k = k / 2; 
    } 
 
    // Print the array elements 
    for(i = 0; i < n; i++)
    { 
        Console.Write(a[i] + " "); 
    } 
}
     
// Driver code
public static void Main(String[] args) 
{
     
    // Given array []arr 
    int []arr = { 5, 20, 30, 40, 36, 
                  33, 25, 15, 10 }; 
    int n = arr.Length; 
     
    // Function call 
    sortArr(arr, n); 
}
}
 
// This code is contributed by gauravrajput1

Output:

Time Complexity: O(N), where N denotes the number of nodes.
Auxiliary Space: O(1)

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My Personal Notes

C++

Java

Python3

C#

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