LeetCode 310. Minimum Height Trees

原题链接在这里：https://leetcode.com/problems/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]

题解：

与Course Schedule, Course Schedule II类似。

用BFS based topological sort. 从叶子节点开始放入queue中，最后剩下的一个或者两个就是最中心的点.

这里练习undirected graph的topological sort. 用Map<Integer, Set<Integer>>来表示graph, 一条edge, 两头node都需要加graph中. 

Time Complexity: O(n+e). Space: O(n+e).

AC Java:

class Solution {
    public List<Integer> findMinHeightTrees(int n, int[][] edges) {
        List<Integer> res = new ArrayList<>();
        if(n == 1){
            return Arrays.asList(0);
        }

        if(n < 1 || edges == null || edges.length != n - 1){
            return res;
        }

        List<HashSet<Integer>> graph = new ArrayList<>();
        for(int i = 0; i < n; i++){
            graph.add(new HashSet<>());
        }

        for(int[] e : edges){
            graph.get(e[0]).add(e[1]);
            graph.get(e[1]).add(e[0]);
        }

        LinkedList<Integer> que = new LinkedList<>();
        for(int i = 0; i < n; i++){
            if(graph.get(i).size() == 1){
                que.add(i);
            }
        }

        while(n > 2){
            n -= que.size();
            LinkedList<Integer> temp = new LinkedList<>();
            while(!que.isEmpty()){
                int cur = que.poll();
                for(int nei : graph.get(cur)){
                    graph.get(cur).remove(nei);
                    graph.get(nei).remove(cur);
                    if(graph.get(nei).size() == 1){
                        temp.add(nei);
                    }
                }
            }

            que = temp;
        }

        return que;
    }
}