计算给定树中权重为偶数的节点

给定一棵树，以及所有节点的权重，任务是计算权重为偶数的节点的数量。

例子：

Input:

Output: 3
Only the weights of the nodes 2, 4 and 5 are even.

方法：在树上执行dfs，对于每个节点，检查其权重是否可被2整除。如果是，则增加计数。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
int ans = 0;
  
vector graph[100];
vector 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;
}

Java

// Java implementation of the approach 
import java.util.*; 
  
class GFG 
{ 
    static int ans = 0; 
      
    @SuppressWarnings("unchecked") 
    static Vector[] 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

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

C#

// C# implementation of the approach 
using System;
using System.Collections.Generic;
  
class GFG 
{ 
    static int ans = 0; 
    static List[] graph = new List[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(); 
           
        // 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

输出：

3

复杂度分析：

• 时间复杂度： O(N)。
  在dfs中，树的每个节点都处理一次，因此，如果树中总共有N个节点，则由于dfs而导致的复杂度为O(N)。因此，时间复杂度为O(N)。
• 辅助空间： O(1)。
  不需要任何额外的空间，因此空间复杂度是恒定的。