在给定的树中找到最大偶数和节点
给定一个具有所有节点权重的树，任务是找到权重为偶数和的最大加权节点。
示例:
Input: Tree =
5
/ \
10 6
/ \
11 8
Output: 11
Explanation:
The tree node weights are:
5 -> 5
10 -> 1 + 0 = 1
6 -> 6
11 -> 1 + 1 = 2
8 -> 8
Here, digit sum for nodes
containing 11, 6 and 8 are even.
Hence, the maximum weighing
even digit sum node is 11.
Input: Tree =
1
/ \
4 7
/ \ / \
11 3 15 8
Output: 15
Explanation:
Here, digit sum for nodes containing
4, 11, 15 and 8 are even.
Hence, the maximum weighing
even digit sum node is 15.
方法:要解决上述问题，请遵循以下步骤:
- 这个想法是对树和每个节点执行深度优先搜索。
- 首先，通过迭代每个数字找到节点中权重的数字和，然后检查节点是否有偶数数字和。
- 最后，如果它有一个偶数和，那么检查这个节点是否是我们到目前为止找到的最大偶数和节点，如果是，那么使这个节点成为最大偶数和节点。
下面是上述方法的实现:
C++
// C++ program to find the
// maximum weighing node
// in the tree whose weight
// has an Even Digit Sum
#include <bits/stdc++.h>
using namespace std;
const int sz = 1e5;
int ans = -1;
vector<int> graph[100];
vector<int> weight(100);
// Function to find the digit sum
// for a number
int digitSum(int num)
{
int sum = 0;
while (num)
{
sum += (num % 10);
num /= 10;
}
return sum;
}
// Function to perform dfs
void dfs(int node, int parent)
{
// Check if the digit sum
// of the node weight
// is even or not
if (!(digitSum(weight[node]) & 1))
ans = max(ans, weight[node]);
// Performing DFS to iterate the
// remaining nodes
for (int to : graph[node]) {
if (to == parent)
continue;
dfs(to, node);
}
}
// Driver code
int main()
{
// 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);
// Call the dfs function to
// traverse through the tree
dfs(1, 1);
cout << ans << endl;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find the
// maximum weighing node
// in the tree whose weight
// has an Even Digit Sum
import java.util.*;
class GFG
{
static int sz = (int)1e5;
static int ans = -1;
static Vector<Integer>[] graph = new Vector[100];
static int[] weight = new int[100];
// Function to find the digit sum
// for a number
static int digitSum(int num)
{
int sum = 0;
while (num > 0)
{
sum += (num % 10);
num /= 10;
}
return sum;
}
// Function to perform dfs
static void dfs(int node, int parent)
{
// Check if the digit sum
// of the node weight
// is even or not
if (!(digitSum(weight[node]) % 2 == 1))
ans = Math.max(ans, weight[node]);
// Performing DFS to iterate the
// remaining nodes
for (int to : graph[node])
{
if (to == parent)
continue;
dfs(to, node);
}
}
// Driver code
public static void main(String[] args)
{
// Weights of the node
weight[1] = 5;
weight[2] = 10;
weight[3] = 11;
weight[4] = 8;
weight[5] = 6;
for (int i = 0; i < graph.length; i++)
graph[i] = new Vector<Integer>();
// Edges of the tree
graph[1].add(2);
graph[2].add(3);
graph[2].add(4);
graph[1].add(5);
// Call the dfs function to
// traverse through the tree
dfs(1, 1);
System.out.print(ans + "\n");
}
}
// This code contributed by Princi Singh
Python 3
# Python3 program to find the
# maximum weighing node
# in the tree whose weight
# has an Even Digit Sum
sz = 1e5
ans = -1
graph = [[] for i in range(100)]
weight = [-1] * 100
# Function to find the digit sum
# for a number
def digitSum(num):
sum = 0
while num:
sum += num % 10
num = num // 10
return sum
# Function to perform dfs
def dfs(node, parent):
global ans
# Check if the digit sum
# of the node weight
# is even or not
if not(digitSum(weight[node]) & 1):
ans = max(ans, weight[node])
# Performing DFS to iterate the
# remaining nodes
for to in graph[node]:
if to == parent:
continue
dfs(to, node)
# Driver code
def main():
# 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)
# Call the dfs function to
# traverse through the tree
dfs(1, 1)
print(ans)
main()
# This code is contributed by stutipathak31jan
C
// C# program to find the
// maximum weighing node
// in the tree whose weight
// has an Even Digit Sum
using System;
using System.Collections.Generic;
class GFG
{
static int sz = (int)1e5;
static int ans = -1;
static List<int>[] graph = new List<int>[ 100 ];
static int[] weight = new int[100];
// Function to find the digit sum
// for a number
static int digitSum(int num)
{
int sum = 0;
while (num > 0)
{
sum += (num % 10);
num /= 10;
}
return sum;
}
// Function to perform dfs
static void dfs(int node, int parent)
{
// Check if the digit sum
// of the node weight
// is even or not
if (!(digitSum(weight[node]) % 2 == 1))
ans = Math.Max(ans, weight[node]);
// Performing DFS to iterate the
// remaining nodes
foreach(int to in graph[node])
{
if (to == parent)
continue;
dfs(to, node);
}
}
// Driver code
public static void Main(String[] args)
{
// Weights of the node
weight[1] = 5;
weight[2] = 10;
weight[3] = 11;
weight[4] = 8;
weight[5] = 6;
for (int i = 0; i < graph.Length; i++)
graph[i] = new List<int>();
// Edges of the tree
graph[1].Add(2);
graph[2].Add(3);
graph[2].Add(4);
graph[1].Add(5);
// Call the dfs function to
// traverse through the tree
dfs(1, 1);
Console.Write(ans + "\n");
}
}
// This code is contributed by Rajput-Ji
java 描述语言
<script>
// Javascript program to find the
// maximum weighing node
// in the tree whose weight
// has an Even Digit Sum
let sz = 1e5;
let ans = -1;
let graph = new Array(100);
let weight = new Array(100);
// Function to find the digit sum
// for a number
function digitSum(num)
{
let sum = 0;
while (num > 0)
{
sum += (num % 10);
num = parseInt(num / 10, 10);
}
return sum;
}
// Function to perform dfs
function dfs(node, parent)
{
// Check if the digit sum
// of the node weight
// is even or not
if (!(digitSum(weight[node]) % 2 == 1))
ans = Math.max(ans, weight[node]);
// Performing DFS to iterate the
// remaining nodes
for (let to = 0; to < graph[node].length; to++)
{
if (graph[node][to] == parent)
continue;
dfs(graph[node][to], node);
}
}
// Weights of the node
weight[1] = 5;
weight[2] = 10;
weight[3] = 11;
weight[4] = 8;
weight[5] = 6;
for (let i = 0; i < graph.length; i++)
graph[i] = [];
// Edges of the tree
graph[1].push(2);
graph[2].push(3);
graph[2].push(4);
graph[1].push(5);
// Call the dfs function to
// traverse through the tree
dfs(1, 1);
document.write(ans + "</br>");
// This code is contributed by decode2207.
</script>
Output
11
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，因此如果树中总共有 N 个节点，由于 dfs 而导致的复杂性是 O(N)。此外，为了处理每个节点，使用了复杂度为 O(d)的 DigitSum()函数，其中 d 是节点权重中的位数，但是由于任何节点的权重都是整数，因此最多只能有 10 位，因此该函数不会影响整体时间复杂度。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，所以空间复杂度不变。
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