310. Minimum Height Trees 难度：medium 类别：图-优快云博客

本文链接：https://blog.youkuaiyun.com/luoyingmin/article/details/52701429

本文介绍了一种算法，用于找到给定树形图中使得树高度最小的根节点，并提供了具体的实现代码。通过迭代删除叶子节点的方式，最终确定满足条件的根节点。

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题目：

For a undirected graph with tree characteristics, we can choose any node as the root. The result graph is then a rooted tree. Among all possible rooted trees, those with minimum height are called minimum height trees (MHTs). Given such a graph, write a function to find all the MHTs and return a list of their root labels.

Format
The graph contains n nodes which are labeled from 0 to n - 1. You will be given the number n and a list of undirected edges (each edge is a pair of labels).

You can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together in edges.

Example 1:

Given n = 4, edges = [[1, 0], [1, 2], [1, 3]]

return [1]

Example 2:

Given n = 6, edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]

return

[3,
 4]

思路：

这题要求我们对于一棵树中，找出一个结点，让它成为根的时候，整个树的高度最小，返回所有可能成为根的结点序列。为了解决这个问题，有一个思路是在每次迭代中，将野子节点删去，直到最后可能剩下一个结点或者两个结点，这一个或两个结点则是我们所要求的结点。可以试着这样去理解这个过程，所求的根结点就是到叶子结点的距离最大值中最小的那个点，假设一开始同时从每个叶子出发向中心靠拢，每走一步就删掉原来的结点，这样就逐步向中心靠拢，在最小的步数内能同时汇聚在一个或两个结点上，这就是我们要求的结果。

程序：

class Solution {
public:
    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
 		vector<vector<int> > adj(n);
	vector<int> degree(n,0);
	set<int> s;
	vector<int> res;
	
	for(int i = 0;i < n;i++)
		s.insert(i);
		
	for(int i = 0;i < edges.size();i++)
	{
		adj[edges[i].first].push_back(edges[i].second);
		adj[edges[i].second].push_back(edges[i].first);
		degree[edges[i].first]++;
		degree[edges[i].second]++;
	}
	
	while(n > 2)
	{
		set<int> temp;
		set<int>::iterator p;
		for(int i = 0;i < degree.size();i++)
		{
			if(degree[i] == 1)
			{
				temp.insert(i);
				s.erase(i);
				n--;
				degree[i]--;
			}
		}
		
		for(p = temp.begin();p != temp.end();p++)
		{
			int u = *p;
			for(int i = 0;i < adj[u].size();i++)
				degree[adj[u][i]]--;
		}
	}
	
	set<int>::iterator p = s.begin();
	for(;p != s.end();p++)
		res.push_back(*p);
	
	return res;
    }
};