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.

思路二 ： 

除了 DP 方案，没有想到其他思路，在网上借鉴了其他了的想法，理解后实现通过。

这个思路实际上是一个 BFS 思路。和常见的从根节点进行 BFS 不同，这里从叶子节点开始进行 BFS。

所有入度（即相连边数）为 1 的节点即是叶子节点。找高度最小的节点，即找离所有叶子节点最远的节点，也即找最中心的节点。

找最中心的节点的思路很简单：

• 每次去掉当前图的所有叶子节点，重复此操作直到只剩下最后的根。

根据这个思路可以回答题目中的 [ hint : How many MHTs can a graph have at most? ]，只能有一个或者两个最小高度树树根。证明省略。

class Solution {public:    class TreeNode{        public:         int val;        unordered_set<TreeNode*> neighber;        TreeNode (int val){            this->val=val;        }            };    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {                map<int,TreeNode*> mp;        for(int i=0;i<n;i++){            TreeNode *p=new TreeNode(i);            mp[i]=p;        }        pair<int,int> pa;        for(int i=0;i<edges.size();i++){            pa=edges[i];            mp[pa.first]->neighber.insert(mp[pa.second]);            mp[pa.second]->neighber.insert(mp[pa.first]);                    }                map<int,TreeNode*>::iterator m_iter;                 while(mp.size()>2){                        vector<TreeNode*> vec;                        for(m_iter=mp.begin();m_iter!=mp.end();m_iter++){                if(m_iter->second->neighber.size()==1){                    vec.push_back(m_iter->second);                                    }                            }                        for(int i=0;i<vec.size();i++){                TreeNode* p=*((vec[i])->neighber.begin());                                p->neighber.erase((vec[i]));                (vec[i])->neighber.erase(p);                                mp.erase((vec[i])->val);                             }                     }                  vector<int> res;                 for(m_iter=mp.begin();m_iter!=mp.end();m_iter++){            res.push_back(m_iter->first);        }        return res;    }};