Leetcode 310. Minimum Height Trees【medium】

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 undirectededges (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 inedges.

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 byexactly 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.

题解：

/*#include <iostream>#include<set>#include<vector>#include<map>#include<list>#include<queue>using namespace std;*/vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {    if(n==1) return vector<int>{0};    map<int,set<int> > flights;    vector<int> buffer1;    vector<int> buffer2;    vector<int>* pb1 = &buffer1;    vector<int>* pb2 = &buffer2;    for(auto& edge:edges){        flights[edge.first].insert(edge.second);        flights[edge.second].insert(edge.first);    }    for(auto& flight:flights){        if(flight.second.size() == 1)            pb2->push_back(flight.first);    }    do{        pb1->clear();        swap(pb1,pb2);        for(auto& b:*pb1){            for(auto& n:flights[b]){                flights[n].erase(b);                if(flights[n].size()==1){                pb2->push_back(n);                }            }        }    }while(!pb2->empty());    return *pb1;}/*int main(){    vector<pair<int, int>> tree0 = {{0,1}};    vector<int> result = findMinHeightTrees(1,tree0);    for(auto& num:result){        cout << num << endl;    }    return 0;}*/

思路：

  第一次，我的想法是对每一个结点得到它的最远路长度，对于最远路长度节点得到它的最远路径，路径的中间就是所求根

  然而到999个节点的案例时就超时了

  第二次，我上网找了下寻找最长路径，也就是图的直径的方法，得到的算法是：对任意一个节点，寻找离他最远的节点，这个节点就是图直径的端点，再从这个端点寻找离他最远的点，就能得到另一个端点，从而得到直径路径

  然而到5000个节点的案例时就超时了

  第三次，我看leetcode上的discuss，vote最高的几个答案都是说：删除所有叶子，然后第二轮再删除所有新的叶子，如此循环直到点集合为空，则最近一轮的叶子就是解，原解：https://discuss.leetcode.com/topic/30535/c-solution-o-n-time-o-n-space/3

  这...怎么想得出来啊，以后还是先在心里想定一个算法，再在discuss里对答案，最后才下手写代码好了，耗不起这时间