LeetCode *** 310. Minimum Height Trees

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]

Hint:

Show Hint

1. How many MHTs can a graph have at most?

分析：

。。

代码：

class Solution {public:    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {        vector<set<int>> record(n);        for(auto edge:edges){            record[edge.first].insert(edge.second);            record[edge.second].insert(edge.first);        }        queue<int> que1;        queue<int> que2;        for(int i=0;i<n;++i)            if(record[i].size()<=1)que1.push(i);        if(n>2){            while(!que1.empty()){                int t=que1.front();                que1.pop();                n--;                            set<int>::iterator it=record[t].begin();                while(it!=record[t].end()){                    record[*it].erase(t);                    if(record[*it].size()==1)que2.push(*it);                    it++;                }                if(n>2&&que1.empty()){                    que1.swap(que2);                }            }            que1.swap(que2);        }            vector<int> res;        while(!que1.empty()){            res.push_back(que1.front());            que1.pop();        }        return res;                    }};