【LeetCode】310. Minimum Height Trees

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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 :

Input: n = 4, edges = [[1, 0], [1, 2], [1, 3]]

        0
        |
        1
       / \
      2   3 

Output: [1]

Example 2 :

Input: n = 6, edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]

     0  1  2
      \ | /
        3
        |
        4
        |
        5 

Output: [3, 4]

链接： http://leetcode.com/problems/minimum-height-trees/

题解：可知最后解只能是一个或两个点，如果三个点以上，以其中一个点为根，其他两点必定在其子树上，因为算法是一层一层的去掉外面的叶节点，所以最后剩下三个点时外面两个点肯定要去掉，因为一层一层的剥开保证了对每个根节点每次都减掉一层也就是最后的根节点来说高度不变这样最后只能有一个或两个

class Solution {
public:
    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
        vector<int> ans;
        if (n == 1) return {0};
        vector<unordered_set<int>> tree(n);
     
      
        for(int i=0;i<edges.size();i++){
            tree[edges[i].first].insert(edges[i].second);
            tree[edges[i].second].insert(edges[i].first);
        }
       
        queue<int> q;
        for(int i=0;i<n;i++){
            if(tree[i].size()==1) q.push(i);
        }
        while(n>2){
            int size=q.size();
            n-=size;
            for(int i=0;i<size;i++){
                int x=q.front();
                q.pop();
                for(auto t:tree[x]){
                    tree[t].erase(x);
                    if(tree[t].size()==1) q.push(t);
                }
            }
        }
        while(!q.empty()){
            ans.push_back(q.front());
            q.pop();
            
        }
        return ans;
    }
    
    
};