计算给定树中权重为
的节点
原文： https://www.geeksforgeeks.org/count-the-nodes-in-the-given-tree-whose-weight-is-even/
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数。
示例：
输入：
输出：3 仅节点 2、4 和 5 的权重相等。
输出：3
仅节点 2、4 和 5 的权重相等。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 <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;
}
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
Python
# 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<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
Output:
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。
如果您喜欢 GeeksforGeeks 并希望做出贡献，则还可以使用 tribution.geeksforgeeks.org 撰写文章，或将您的文章邮寄至 tribution@geeksforgeeks.org。 查看您的文章出现在 GeeksforGeeks 主页上，并帮助其他 Geeks。
如果您发现任何不正确的地方，请单击下面的“改进文章”按钮，以改进本文。
版权属于：月萌API www.moonapi.com，转载请注明出处
