Count the nodes in the given tree whose weight is even
Given a tree, and the weights of all the nodes, the task is to count the number of nodes whose weight is even.
Examples:
Input:
Output: 3
Only the weights of the nodes 2, 4 and 5 are even.
Approach: Perform dfs on the tree and for every node, check if it’s weight is divisible by 2 or not. If yes then increment count.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; int ans = 0; vector< int > graph[100]; vector< int > weight(100); // Function to perform dfs void dfs( int node, int parent) { // If weight of the current node is even if (weight[node] % 2 == 0) ans += 1; for ( int to : graph[node]) { if (to == parent) continue ; dfs(to, node); } } // Driver code int main() { int x = 15; // Weights of the node weight[1] = 5; weight[2] = 10; weight[3] = 11; weight[4] = 8; weight[5] = 6; // Edges of the tree graph[1].push_back(2); graph[2].push_back(3); graph[2].push_back(4); graph[1].push_back(5); dfs(1, 1); cout << ans; return 0; } |
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Java
// Java implementation of the approach import java.util.*; class GFG { static int ans = 0 ; @SuppressWarnings ( "unchecked" ) static Vector<Integer>[] graph = new Vector[ 100 ]; static int [] weight = new int [ 100 ]; // Function to perform dfs static void dfs( int node, int parent) { // If weight of the current node is even if (weight[node] % 2 == 0 ) ans += 1 ; for ( int to : graph[node]) { if (to == parent) continue ; dfs(to, node); } } // Driver code public static void main(String[] args) { int x = 15 ; for ( int i = 0 ; i < 100 ; i++) graph[i] = new Vector<>(); // Weights of the node weight[ 1 ] = 5 ; weight[ 2 ] = 10 ; weight[ 3 ] = 11 ; weight[ 4 ] = 8 ; weight[ 5 ] = 6 ; // Edges of the tree graph[ 1 ].add( 2 ); graph[ 2 ].add( 3 ); graph[ 2 ].add( 4 ); graph[ 1 ].add( 5 ); dfs( 1 , 1 ); System.out.println(ans); } } // This code is contributed by shubhamsingh10 |
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Python3
# Python3 implementation of the approach ans = 0 graph = [[] for i in range ( 100 )] weight = [ 0 ] * 100 # Function to perform dfs def dfs(node, parent): global ans # If weight of the current node is even if (weight[node] % 2 = = 0 ): ans + = 1 for to in graph[node]: if (to = = parent): continue dfs(to, node) # Driver code x = 15 # Weights of the node weight[ 1 ] = 5 weight[ 2 ] = 10 weight[ 3 ] = 11 weight[ 4 ] = 8 weight[ 5 ] = 6 # Edges of the tree graph[ 1 ].append( 2 ) graph[ 2 ].append( 3 ) graph[ 2 ].append( 4 ) graph[ 1 ].append( 5 ) dfs( 1 , 1 ) print (ans) # This code is contributed by SHUBHAMSINGH10 |
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C#
// C# implementation of the approach using System; using System.Collections.Generic; class GFG { static int ans = 0; static List< int >[] graph = new List< int >[100]; static int [] weight = new int [100]; // Function to perform dfs static void dfs( int node, int parent) { // If weight of the current node is even if (weight[node] % 2 == 0) ans += 1; foreach ( int to in graph[node]) { if (to == parent) continue ; dfs(to, node); } } // Driver code public static void Main(String[] args) { for ( int i = 0; i < 100; i++) graph[i] = new List< int >(); // Weights of the node weight[1] = 5; weight[2] = 10; weight[3] = 11; weight[4] = 8; weight[5] = 6; // Edges of the tree graph[1].Add(2); graph[2].Add(3); graph[2].Add(4); graph[1].Add(5); dfs(1, 1); Console.WriteLine(ans); } } // This code is contributed by Rajput-Ji |
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Output:
3
Complexity Analysis:
- Time Complexity : O(N).
In dfs, every node of the tree is processed once and hence the complexity due to the dfs is O(N) if there are total N nodes in the tree. Therefore, the time complexity is O(N). - Auxiliary Space : O(1).
Any extra space is not required, so the space complexity is constant.
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