310. Minimum Height Trees

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.

题意：
一个n个结点n-1条边的无向无环图（也就是树），找出以哪个节点为root时，树的高度最小？（返回所有答案）

思路：
最开始我的思路是递归，类似于DFS：先遍历得到每个结点为root时得到的树的高度，然后排序找到最小值。但写出的程序runtime error。一时找不到bug。

从网上借鉴一个很巧妙的思路：类似于“剥洋葱”，一层一层的删除叶节点（入度为1），这样最后得到的一个或两个入度为1的结点就是所求。最后答案的节点数不可能大于2。假设答案有3个结点，则分别以这三个结点为root时，形成树的高度相同。用反证法证明：（觉得手稿会更清楚）

class Solution {
public:
    vector<int> findMinHeightTrees(int n, vector<pair<int, int> >& edges) {
        if (n == 1) return {0}; //注意特例的处理，如果只有一个结点0，下面的算法不适用，直接返回{0}
        vector<int> res, d(n, 0);//res保存答案，d保存每个结点的入度
        vector<vector<int> > g(n, vector<int>()); //g保存每个结点的相邻结点
        queue<int> q; //q为一个用来保存入度为1的结点的队列
        for (auto a : edges) { //依次处理每条边，初始化d和g
            g[a.first].push_back(a.second);
            ++d[a.first];
            g[a.second].push_back(a.first);
            ++d[a.second];
        }
        for (int i = 0; i < n; ++i) {//将初始时入度为1的结点保存到q中
            if (d[i] == 1) q.push(i);
        }
        while (n > 2) {如果<=2，说明达到最后一轮，跳出
            int sz = q.size();
            for (int i = 0; i < sz; ++i) { \\依次处理每轮中入度为1的结点，处理一个，pop一个
                int t = q.front(); q.pop();
                --n;
                for (int i : g[t]) { \\将入度为1的点的相邻结点入度全部-1，如果得到新的入度为1的点，则保存到q中
                    --d[i];
                    if (d[i] == 1) q.push(i);
                }
            }
        }
        while (!q.empty()) {
            res.push_back(q.front()); q.pop();
        }
        return res;
    }
};