[leetcode]310. Minimum Height Trees

题目链接：https://leetcode.com/problems/minimum-height-trees/tabs/description

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]]

        0        |        1       / \      2   3

return [1]

Example 2:

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

     0  1  2      \ | /        3        |        4        |        5

return [3, 4]

Note:

(1) According to the definition of tree on Wikipedia: “a tree is an undirected graph in which any two vertices are connected by exactly one path. In other words, any connected graph without simple cycles is a tree.”

(2) The height of a rooted tree is the number of edges on the longest downward path between the root and a leaf.

思路一：

这个思路实际上是一个 BFS 思路。和常见的从根节点进行 BFS 不同，这里从叶子节点开始进行 BFS。

所有入度（即相连边数）为 1 的节点即是叶子节点。找高度最小的节点，即找离所有叶子节点最远的节点，也即找最中心的节点。

找最中心的节点的思路很简单：

• 每次去掉当前图的所有叶子节点，重复此操作直到只剩下最后的根。

class Solution {public:    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {        vector<unordered_set<int>> adj(n);        for(pair<int,int>p:edges)        {            adj[p.first].insert(p.second);            adj[p.second].insert(p.first);        }        vector<int> current;        if(n==1)        {            current.push_back(0);            return current;        }        for(int i=0;i<adj.size();i++)        {            if(adj[i].size()==1)            {                current.push_back(i);            }        }        while(true)        {            vector<int> next;            for(int node:current)            {                for(int neighbor:adj[node])                {                    adj[neighbor].erase(node);                    if(adj[neighbor].size()==1)                        next.push_back(neighbor);                }            }            if(next.empty())                return current;            current=next;        }    }};

思路二：