[leetcode]310. Minimum Height Trees-优快云博客

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]]

return [1]

Example 2:

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

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：