310. Minimum Height Trees（最小高度树）

题目链接

https://leetcode.com/problems/minimum-height-trees/description/

题目描述

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.

Example1:
Given n = 4, edges = [[1, 0], [1, 2], [1, 3]

  0  |  1 / \2   3

return [1]

Example2:
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.

题目分析

这道题要求树具有最小高度，考虑什么情况下树的高度最小。经过分析可以发现：当以图中的最长链的中点作为树的根时，这颗树的高度最小。那么接下来的问题就是如何找到最长链的中点。
假设我们已经找到这样的一颗树，它的高度最小。不断地删除它的叶子节点（度为1）那么最后剩下的1或2（最长链长度为偶数时）个节点就是所求答案。

方法：DFS

算法描述

1. 以某种方式储存图，同时记录每个节点的度
2. 删除度为1的节点，并把与之相连的节点度数减1
3. 重复步骤2直到剩下1或2个节点

参考代码

class Solution {public:    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {        map<int, set<int> > graph;        graph.insert(make_pair(0, set<int>()));        for (auto e : edges) {            if (graph.find(e.first) == graph.end())                graph.insert(make_pair(e.first, set<int>()));            graph[e.first].insert(e.second);            if (graph.find(e.second) == graph.end())                graph.insert(make_pair(e.second, set<int>()));            graph[e.second].insert(e.first);        }        queue<int> leaves, newLeaves;        for (auto node : graph)            if (node.second.size() == 1)                leaves.push(node.first);        while (graph.size() > 2 || (graph.size() == 2 && leaves.size() != 2)) {            int leaf = leaves.front();            leaves.pop();            int neighbor = *(graph[leaf].begin());            graph[neighbor].erase(leaf);            if (graph[neighbor].size() == 1)                newLeaves.push(neighbor);            graph.erase(leaf);            if (leaves.empty()) {                leaves = newLeaves;                newLeaves = queue<int>();            }        }        vector<int> ans;        for (auto node : graph) {            ans.push_back(node.first);        }        return ans;    }};