LeetCode 310 Minimum Height Trees

题目

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]

解法

图中最长路径的中点就最小高度树的根，当最长路径是偶数的时候，有两个最小高度树，为奇数时有一个最小高度树，所以最后结果中元素个数不超过2。图中度数为1的结点可以看做叶子结点，把叶子节点一层层去掉，最后剩下不超过两个的结点就是最小高度树的根。

class Solution {public:    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {        map<int, vector<int>> graph;        vector<int> degree(n, 0);          for (int i = 0; i < edges.size(); i++) {            graph[edges[i].first].push_back(edges[i].second);            graph[edges[i].second].push_back(edges[i].first);            degree[edges[i].first]++;            degree[edges[i].second]++;        }        int count = n;        while(count > 2) {            vector<int> temp;            for(int i = 0; i < n; i++) {                  if(degree[i] == 1)                    temp.push_back(i);            }              for(int j : temp) {                degree[j] = -1;                count--;                for(int k = 0; k < graph[j].size(); k++)                      degree[graph[j][k]]--;             }        }        vector<int> res;        for(int i= 0; i < degree.size(); i++) {            if(degree[i]>=0)                res.push_back(i);        }        return res;     }};