计算给定树中权重总和为奇数的节点

给定一棵树，以及所有节点的权重，任务是计算权重位数之和为奇数的节点数。
例子：

Input: 
 

Output: 3 
Node 1: digitSum(144) = 1 + 4 + 4 = 9 
Node 2: digitSum(1234) = 1 + 2 + 3 + 4 = 10 
Node 3: digitSum(21) = 2 + 1 = 3 
Node 4: digitSum(5) = 5 
Node 5: digitSum(77) = 7 + 7 = 14 
Only the sum of digits of the weights of nodes 1, 3 and 4 are odd. 
 

方法：对树执行 dfs，对于每个节点，检查其权重的数字之和是否为奇数。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 return the
// sum of the digits of n
int digitSum(int n)
{
    int sum = 0;
    while (n) {
        sum += n % 10;
        n = n / 10;
    }
    return sum;
}
 
// Function to perform dfs
void dfs(int node, int parent)
{
    // If sum of the digits of current node's
    // weight is odd then increment ans
    int sum = digitSum(weight[node]);
    if (sum % 2 == 1)
        ans += 1;
 
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
 
// Driver code
int main()
{
 
    // Weights of the node
    weight[1] = 144;
    weight[2] = 1234;
    weight[3] = 21;
    weight[4] = 5;
    weight[5] = 77;
 
    // 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;
 
    static Vector[] graph = new Vector[100];
    static Integer[] weight = new Integer[100];
 
    // Function to return the
    // sum of the digits of n
    static int digitSum(int n)
    {
        int sum = 0;
        while (n > 0)
        {
            sum += n % 10;
            n = n / 10;
        }
        return sum;
    }
 
    // Function to perform dfs
    static void dfs(int node, int parent)
    {
         
        // If sum of the digits of current node's
        // weight is odd then increment ans
        int sum = digitSum(weight[node]);
        if (sum % 2 == 1)
            ans += 1;
 
        for (int to : 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 Vector();
         
        // Weights of the node
        weight[1] = 144;
        weight[2] = 1234;
        weight[3] = 21;
        weight[4] = 5;
        weight[5] = 77;
 
        // 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.print(ans);
    }
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach
ans = 0
 
graph = [[] for i in range(100)]
weight = [0] * 100
 
# Function to return the
# sum of the digits of n
def digitSum(n):
    sum = 0
    while (n):
        sum += n % 10
        n = n // 10
    return sum
 
# Function to perform dfs
def dfs(node, parent):
    global ans
 
    # If sum of the digits of current node's
    # weight is odd then increment ans
    sum = digitSum(weight[node])
    if (sum % 2 == 1):
        ans += 1
     
    for to in graph[node]:
        if (to == parent):
            continue
        dfs(to, node)
 
# Driver code
 
# Weights of the node
weight[1] = 144
weight[2] = 1234
weight[3] = 21
weight[4] = 5
weight[5] = 77
 
# 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 return the
    // sum of the digits of n
    static int digitSum(int n)
    {
        int sum = 0;
        while (n > 0)
        {
            sum += n % 10;
            n = n / 10;
        }
        return sum;
    }
 
    // Function to perform dfs
    static void dfs(int node, int parent)
    {
         
        // If sum of the digits of current node's
        // weight is odd then increment ans
        int sum = digitSum(weight[node]);
        if (sum % 2 == 1)
            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] = 144;
        weight[2] = 1234;
        weight[3] = 21;
        weight[4] = 5;
        weight[5] = 77;
 
        // Edges of the tree
        graph[1].Add(2);
        graph[2].Add(3);
        graph[2].Add(4);
        graph[1].Add(5);
 
        dfs(1, 1);
 
        Console.Write(ans);
    }
}
 
// This code is contributed by PrinciRaj1992

Javascript

 
// JavaScript implementation of the approach
var ans = 0;
var graph = Array.from(Array(100), ()=>Array());
var weight = Array(100);
// Function to return the
// sum of the digits of n
function digitSum(n)
{
    var sum = 0;
    while (n > 0)
    {
        sum += n % 10;
        n = parseInt( n / 10);
    }
    return sum;
}
// Function to perform dfs
function dfs(node, parent)
{
     
    // If sum of the digits of current node's
    // weight is odd then increment ans
    var sum = digitSum(weight[node]);
    if (sum % 2 == 1)
        ans += 1;
    for(var to of graph[node])
    {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
// Driver code
for (var i = 0; i < 100; i++)
    graph[i] = [];
 
// Weights of the node
weight[1] = 144;
weight[2] = 1234;
weight[3] = 21;
weight[4] = 5;
weight[5] = 77;
// Edges of the tree
graph[1].push(2);
graph[2].push(3);
graph[2].push(4);
graph[1].push(5);
dfs(1, 1);
document.write(ans);
 
 

输出：

3

复杂度分析：

• 时间复杂度： O(N)。
  在 DFS 中，树的每个节点都被处理一次，因此对于树中的 N 个节点，由于 dfs 的复杂性是 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。

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