Problem Description

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]

Some Details

• 樹的性質

Solution

極其僵化的思路是想辦法dp統計每個點為根的深度。
比較僵化的思路是最長鏈中間的點。
靈性思路是不斷找到樹的葉子，刪除，更新，最後剩餘的小於等於兩個節點一定是可行的根。
然而僵化的我沒有立刻想到，結束。

Code

vector<int> f[10010];
int len[10010];
bool pd[10010];
class Solution {
public:
    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
        int a,b,i,j,k;
        vector<int>ans;
        if (n==0) return ans;
        for (i=0;i<n;i++) {len[i]=0;f[i].clear();}
        for (i=0;i<edges.size();i++)
        {
            a=edges[i].first; b=edges[i].second;
            len[a]++; len[b]++;
            f[a].push_back(b); f[b].push_back(a);
        }
        for (i=0;i<n;i++) pd[i]=false;
        k=n;
        vector<int>now;
        while (k>2)
        {
            now.clear();
            for (i=0;i<n;i++)
                if (!pd[i] && len[i]==1) now.push_back(i);
            k-=now.size();
            for (i=0;i<now.size();i++)
            {
                a=now[i]; pd[a]=true;
                for (j=0;j<f[a].size();j++) len[f[a][j]]--;
            }
        }
        for (i=0;i<n;i++)
            if (!pd[i]) ans.push_back(i);
        return ans;
    }
};