给定一个由值为[1, N]的节点和一个数组value[]组成的N 元树，其中每个节点i与value[i]相关联，任务是找到所有级别的所有节点值之和的最大值N 叉树。
例子：
Input: N = 8, Edges[][2] = {{0, 1}, {0, 2}, {0, 3}, {1, 4}, {1, 5}, {3, 6}, {6, 7}}, Value[] = {4, 2, 3, -5, -1, 3, -2, 6}
Output: 6
Explanation:
Sum of all nodes of 0th level is 4
Sum of all nodes of 1st level is 0
Sum of all the nodes of 3rd level is 0.
Sum of all the odes of 4th level is 6.
Therefore, maximum sum of any level of the tree is 6.
Input: N = 10, Edges[][2] = {{0, 1}, {0, 2}, {0, 3}, {1, 4}, {1, 5}, {3, 6}, {6, 7}, {6, 8}, {6, 9}}, Value[] = {1, 2, -1, 3, 4, 5, 8, 6, 12, 7}
Output: 25
方法：这个问题可以使用Level order Traversal来解决。在进行遍历时，分别处理不同层次的节点。对于正在处理的每个级别，计算该级别的节点总和并跟踪最大总和。按照步骤：
- 将当前级别的所有子节点存储到队列中，然后在特定级别的级别顺序遍历完成后计算当前级别的节点总和。
- 由于队列现在包含下一层的所有节点，因此可以通过遍历队列轻松计算下一层的节点总数。
- 对连续级别执行相同的过程并更新在每个级别找到的最大节点总和。
- 完成上述步骤后，打印存储的值的最大总和。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the maximum sum
// a level in N-ary treeusing BFS
int maxLevelSum(int N, int M,
vector Value,
int Edges[][2])
{
// Stores the edges of the graph
vector adj[N];
// Create Adjacency list
for (int i = 0; i < M; i++) {
adj[Edges[i][0]].push_back(
Edges[i][1]);
}
// Initialize result
int result = Value[0];
// Stores the nodes of each level
queue q;
// Insert root
q.push(0);
// Perform level order traversal
while (!q.empty()) {
// Count of nodes of the
// current level
int count = q.size();
int sum = 0;
// Traverse the current level
while (count--) {
// Dequeue a node from queue
int temp = q.front();
q.pop();
// Update sum of current level
sum = sum + Value[temp];
// Enqueue the children of
// dequeued node
for (int i = 0;
i < adj[temp].size(); i++) {
q.push(adj[temp][i]);
}
}
// Update maximum level sum
result = max(sum, result);
}
// Return the result
return result;
}
// Driver Code
int main()
{
// Number of nodes
int N = 10;
// Edges of the N-ary tree
int Edges[][2] = { { 0, 1 }, { 0, 2 },
{ 0, 3 }, { 1, 4 },
{ 1, 5 }, { 3, 6 },
{ 6, 7 }, { 6, 8 },
{ 6, 9 } };
// Given cost
vector Value = { 1, 2, -1, 3, 4,
5, 8, 6, 12, 7 };
// Function call
cout << maxLevelSum(N, N - 1,
Value, Edges);
return 0;
} Java
// Java program for the above approach
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Queue;
class GFG{
// Function to find the maximum sum
// a level in N-ary treeusing BFS
@SuppressWarnings("unchecked")
static int maxLevelSum(int N, int M, int[] Value,
int Edges[][])
{
// Stores the edges of the graph
ArrayList[] adj = new ArrayList[N];
for(int i = 0; i < N; i++)
{
adj[i] = new ArrayList<>();
}
// Create Adjacency list
for(int i = 0; i < M; i++)
{
adj[Edges[i][0]].add(Edges[i][1]);
}
// Initialize result
int result = Value[0];
// Stores the nodes of each level
Queue q = new LinkedList<>();
// Insert root
q.add(0);
// Perform level order traversal
while (!q.isEmpty())
{
// Count of nodes of the
// current level
int count = q.size();
int sum = 0;
// Traverse the current level
while (count-- > 0)
{
// Dequeue a node from queue
int temp = q.poll();
// Update sum of current level
sum = sum + Value[temp];
// Enqueue the children of
// dequeued node
for(int i = 0; i < adj[temp].size(); i++)
{
q.add(adj[temp].get(i));
}
}
// Update maximum level sum
result = Math.max(sum, result);
}
// Return the result
return result;
}
// Driver Code
public static void main(String[] args)
{
// Number of nodes
int N = 10;
// Edges of the N-ary tree
int[][] Edges = { { 0, 1 }, { 0, 2 },
{ 0, 3 }, { 1, 4 },
{ 1, 5 }, { 3, 6 },
{ 6, 7 }, { 6, 8 },
{ 6, 9 } };
// Given cost
int[] Value = { 1, 2, -1, 3, 4,
5, 8, 6, 12, 7 };
// Function call
System.out.println(maxLevelSum(N, N - 1,
Value, Edges));
}
}
// This code is contributed by sanjeev2552 Python3
# Python3 program for the above approach
from collections import deque
# Function to find the maximum sum
# a level in N-ary treeusing BFS
def maxLevelSum(N, M, Value, Edges):
# Stores the edges of the graph
adj = [[] for i in range(N)]
# Create Adjacency list
for i in range(M):
adj[Edges[i][0]].append(Edges[i][1])
# Initialize result
result = Value[0]
# Stores the nodes of each level
q = deque()
# Insert root
q.append(0)
# Perform level order traversal
while (len(q) > 0):
# Count of nodes of the
# current level
count = len(q)
sum = 0
# Traverse the current level
while (count):
# Dequeue a node from queue
temp = q.popleft()
# Update sum of current level
sum = sum + Value[temp]
# Enqueue the children of
# dequeued node
for i in range(len(adj[temp])):
q.append(adj[temp][i])
count -= 1
# Update maximum level sum
result = max(sum, result)
# Return the result
return result
# Driver Code
if __name__ == '__main__':
# Number of nodes
N = 10
# Edges of the N-ary tree
Edges = [ [ 0, 1 ], [ 0, 2 ],
[ 0, 3 ], [ 1, 4 ],
[ 1, 5 ], [ 3, 6 ],
[ 6, 7 ], [ 6, 8 ],
[ 6, 9 ] ]
# Given cost
Value = [ 1, 2, -1, 3, 4,
5, 8, 6, 12, 7 ]
# Function call
print(maxLevelSum(N, N - 1,
Value, Edges))
# This code is contributed by mohit kumar 29C#
// C# program for the
// above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the maximum sum
// a level in N-ary treeusing BFS
static int maxLevelSum(int N, int M,
int[] Value,
int [,]Edges)
{
// Stores the edges of the graph
List[] adj = new List[N];
for(int i = 0; i < N; i++)
{
adj[i] = new List();
}
// Create Adjacency list
for(int i = 0; i < M; i++)
{
adj[Edges[i, 0]].Add(Edges[i, 1]);
}
// Initialize result
int result = Value[0];
// Stores the nodes of each level
Queue q = new Queue();
// Insert root
q.Enqueue(0);
// Perform level order
// traversal
while (q.Count != 0)
{
// Count of nodes of the
// current level
int count = q.Count;
int sum = 0;
// Traverse the current
// level
while (count-- > 0)
{
// Dequeue a node from
// queue
int temp = q.Peek();
q.Dequeue();
// Update sum of current
// level
sum = sum + Value[temp];
// Enqueue the children of
// dequeued node
for(int i = 0;
i < adj[temp].Count; i++)
{
q.Enqueue(adj[temp][i]);
}
}
// Update maximum level sum
result = Math.Max(sum, result);
}
// Return the result
return result;
}
// Driver Code
public static void Main(String[] args)
{
// Number of nodes
int N = 10;
// Edges of the N-ary tree
int[,] Edges = {{0, 1}, {0, 2},
{0, 3}, {1, 4},
{1, 5}, {3, 6},
{6, 7}, {6, 8},
{6, 9}};
// Given cost
int[] Value = {1, 2, -1, 3, 4,
5, 8, 6, 12, 7};
// Function call
Console.WriteLine(maxLevelSum(N, N - 1,
Value, Edges));
}
}
// This code is contributed by Princi Singh 输出
25时间复杂度： O(N)
辅助空间： O(N)
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