leetcode 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.

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；

class Solution {
public:
    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
        vector<vector<int>> g(n,vector<int>());
        vector<int> d(n,0);
        int len = edges.size();
        for(int i=0;i<len;i++)
        {
            int a = edges[i].first;
            int b = edges[i].second;
            g[a].push_back(b);
            g[b].push_back(a);
            d[a]++;
            d[b]++;
        }
        queue<int> q;
        for(int i=0;i<n;i++)
        {
            if(d[i]==1)
                q.push(i);
        }
        while(n>2)
        {
            int len2 = q.size();
            for(int i=0;i<len2;i++)
            {
                 n--;
                 int t = q.front();
                 q.pop();
                 for(int j=0;j<g[t].size();j++)
                 {
                     d[g[t][j]]--;
                     if(d[g[t][j]]==1)
                        q.push(g[t][j]);
                 }
            }
        }
        vector<int> ans;
        while(!q.empty())
        {
            ans.push_back(q.front());
            q.pop();
        }
        if(ans.size()==0)
            ans.push_back(0);
        return ans;
    }
};