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 : “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：

（最初想法：以任意一个点为根节点，然后广度优先遍历图，找到最小深度，时间复杂度O(N*(N+E))，结果超时）

后参考http://www.cnblogs.com/grandyang/p/5000291.html

从叶节点开始，一层一层深入到中心，最后剩余的节点（小于两个）即为结果集；

注意只有一个点的情况；

1 class Solution { 2 public: 3     vector
      
        findMinHeightTrees(int n, vector
       
        
         >& edges) { 4         vector
         
          
           > g(n,vector
           
            ()); 5 vector
            
              d(n,0); 6 int len = edges.size(); 7 for(int i=0;i
             

q;17 for(int i=0;i

2)23 {24 int len2 = q.size();25 for(int i=0;i

ans;39 while(!q.empty())40 {41 ans.push_back(q.front());42 q.pop();43 }44 if(ans.size()==0)45 ans.push_back(0);46 return ans;47 }48 };