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; } |
chevron_right
filter_none
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 |
chevron_right
filter_none
Python3
# Python3 program to implement # the above approachmaxm = 0# Node classclass Node: # Initialise constructor def __init__(self, x): self.val = x self.left = None self.right = None # Function to count the number of 1 in numberdef 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 leafdef 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 coderoot = 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 |
chevron_right
filter_none
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 codepublic 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 |
chevron_right
filter_none
Output:
12
Time Complexity: O(N), where N denotes the number of nodes.
Auxiliary Space: O(1)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
