给定一棵树，以及所有节点的权重，任务是找到加权和最小的子树的根。
例子：
Input:
Output: 5
Weight of sub-tree for parent 1 = ((-1) + (5) + (-2) + (-1) + (3)) = 4
Weight of sub-tree for parent 2 = ((5) + (-1) + (3)) = 7
Weight of sub-tree for parent 3 = -1
Weight of sub-tree for parent 4 = 3
Weight of sub-tree for parent 5 = -2
Node 5 gives the minimum sub-tree weighted sum.
方法：在树上执行dfs，并为每个节点计算以当前节点为根的子树加权总和，然后找到一个节点的最小总和值。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
int ans = 0, mini = INT_MAX;
vector graph[100];
vector weight(100);
// Function to perform dfs and update the tree
// such that every node's weight is the sum of
// the weights of all the nodes in the sub-tree
// of the current node including itself
void dfs(int node, int parent)
{
for (int to : graph[node]) {
if (to == parent)
continue;
dfs(to, node);
// Calculating the weighted
// sum of the subtree
weight[node] += weight[to];
}
}
// Function to find the node
// having minimum sub-tree sum
void findMin(int n)
{
// For every node
for (int i = 1; i <= n; i++) {
// If current node's weight
// is minimum so far
if (mini > weight[i]) {
mini = weight[i];
ans = i;
}
}
}
// Driver code
int main()
{
int n = 5;
// Weights of the node
weight[1] = -1;
weight[2] = 5;
weight[3] = -1;
weight[4] = 3;
weight[5] = -2;
// 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);
findMin(n);
cout << ans;
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
static int ans = 0, mini = Integer.MAX_VALUE;
@SuppressWarnings("unchecked")
static Vector[] graph = new Vector[100];
static Integer[] weight = new Integer[100];
// Function to perform dfs and update the tree
// such that every node's weight is the sum of
// the weights of all the nodes in the sub-tree
// of the current node including itself
static void dfs(int node, int parent)
{
for (int to : graph[node])
{
if (to == parent)
continue;
dfs(to, node);
// Calculating the weighted
// sum of the subtree
weight[node] += weight[to];
}
}
// Function to find the node
// having minimum sub-tree sum x
static void findMin(int n)
{
// For every node
for (int i = 1; i <= n; i++)
{
// If current node's weight x
// is minimum so far
if (mini > weight[i])
{
mini = weight[i];
ans = i;
}
}
}
// Driver code
public static void main(String[] args)
{
int n = 5;
for (int i = 0; i < 100; i++)
graph[i] = new Vector();
// Weights of the node
weight[1] = -1;
weight[2] = 5;
weight[3] = -1;
weight[4] = 3;
weight[5] = -2;
// Edges of the tree
graph[1].add(2);
graph[2].add(3);
graph[2].add(4);
graph[1].add(5);
dfs(1, 1);
findMin(n);
System.out.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, mini = int.MaxValue;
static List[] graph = new List[100];
static int[] weight = new int[100];
// Function to perform dfs and update the tree
// such that every node's weight is the sum of
// the weights of all the nodes in the sub-tree
// of the current node including itself
static void dfs(int node, int parent)
{
foreach (int to in graph[node])
{
if (to == parent)
continue;
dfs(to, node);
// Calculating the weighted
// sum of the subtree
weight[node] += weight[to];
}
}
// Function to find the node
// having minimum sub-tree sum x
static void findMin(int n)
{
// For every node
for (int i = 1; i <= n; i++)
{
// If current node's weight x
// is minimum so far
if (mini > weight[i])
{
mini = weight[i];
ans = i;
}
}
}
// Driver code
public static void Main(String[] args)
{
int n = 5;
for (int i = 0; i < 100; i++)
graph[i] = new List();
// Weights of the node
weight[1] = -1;
weight[2] = 5;
weight[3] = -1;
weight[4] = 3;
weight[5] = -2;
// Edges of the tree
graph[1].Add(2);
graph[2].Add(3);
graph[2].Add(4);
graph[1].Add(5);
dfs(1, 1);
findMin(n);
Console.Write(ans);
}
}
// This code is contributed by Rajput-Ji Python3
# Python3 implementation of the approach
ans = 0
mini = 2**32
graph = [[] for i in range(100)]
weight = [0]*100
# Function to perform dfs and update the tree
# such that every node's weight is the sum of
# the weights of all the nodes in the sub-tree
# of the current node including itself
def dfs(node, parent):
global mini, graph, weight, ans
for to in graph[node]:
if (to == parent):
continue
dfs(to, node)
# Calculating the weighted
# sum of the subtree
weight[node] += weight[to]
# Function to find the node
# having minimum sub-tree sum
def findMin(n):
global mini, graph, weight, ans
# For every node
for i in range(1, n + 1):
# If current node's weight
# is minimum so far
if (mini > weight[i]):
mini = weight[i]
ans = i
# Driver code
n = 5
# Weights of the node
weight[1] = -1
weight[2] = 5
weight[3] = -1
weight[4] = 3
weight[5] = -2
# Edges of the tree
graph[1].append(2)
graph[2].append(3)
graph[2].append(4)
graph[1].append(5)
dfs(1, 1)
findMin(n)
print(ans)
# This code is contributed by SHUBHAMSINGH10输出：
5
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(n)。
递归堆栈。