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 structurestruct Node { int val; // Pointers to left // and right child Node *left, *right; // Initialize constructor Node(int x) { val = x; left = NULL; right = NULL; }};// Function to find the maximum// count of setbits in a root to leafvoid 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 Codeint 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 approachimport java.util.*;class GFG{ static int maxm = 0;// Node structurestatic class Node{ int val; // Pointers to left // and right child Node left, right; // Initialize constructor Node(int x) { val = x; left = null; right = null; }};// Function to find the maximum// count of setbits in a root to leafstatic 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 Codepublic 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 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 |
C#
// C# program for the above approachusing System;class GFG{ // Function to Sort a Bitonic array// in constant spacestatic void sortArr(int []a, int n){ int i, k; // Initialise 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 |
Javascript
<script>// JavaScript Program to implement// the above approachlet maxm = 0;// Node structureclass Node{ // Initialize constructor constructor(x) { this.val = x; this.left = null; this.right = null; }}var root;// Function to find the maximum// count of setbits in a root to leaffunction maxm_setbits(root, ans){ // Check if root is not null if (!root) return; if (root.left == null && root.right == null) { ans += (root.val).toString(2).split('').filter( y => y == '1').length; // Update the maximum count // of setbits maxm = Math.max(ans, maxm); return; } // Traverse left of binary tree maxm_setbits(root.left, ans + (root.val).toString(2).split('').filter( y => y == '1').length); // Traverse right of the binary tree maxm_setbits(root.right, ans + (root.val).toString(2).split('').filter( y => y == '1').length);}// Driver Coderoot = 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);document.write(maxm);// This code is contributed by Dharanendra L V.</script> |
Output:
12
Time Complexity: O(N), where N denotes the number of nodes.
Auxiliary Space: O(1)
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