leetcode 310 : Minimum Height Trees

1、原题如下：

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.

2、解题如下：

class Solution {
public:
    struct vertex
    {
        unordered_set<int> neighbor;
        bool isLeaf() const{return neighbor.size()==1;}
    };
    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
        vector<int> tmp1;
        vector<int> tmp2;
        vector<int>* tmp1_1=&tmp1;
        vector<int>* tmp2_1=&tmp2;
        if(n==1)
        {
            tmp1.push_back(0);
            return tmp1;
        }
        if(n==2)
        {
            tmp1.push_back(0);
            tmp1.push_back(1);
            return tmp1;
        }
        vector<vertex> vertices(n);
        for(auto i:edges)
        {
            vertices[i.first].neighbor.insert(i.second);
            vertices[i.second].neighbor.insert(i.first);
        }
        for(int i=0;i<n;i++)
        {
            if(vertices[i].isLeaf())
            {
                tmp1_1->push_back(i);
            }
        }
        while(1)
        {
            for(auto j:*tmp1_1)
            {
                for(auto k: vertices[j].neighbor)
                {
                    vertices[k].neighbor.erase(j);
                    if(vertices[k].isLeaf()) tmp2_1->push_back(k);
                }
            }
            if(tmp2_1->empty())
            {
                return *tmp1_1;
            }
            tmp1_1->clear();
            swap(tmp1_1,tmp2_1);
        }
    }
};