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]

一道图的题目，我们可以每次都删除叶子节点，维护一个入度为1的队列，删除叶子节点的同时，将新的入度为0的及诶单更新到队列中，直到最后剩下一个或两个节点。代码如下：

public class Solution {
    public List<Integer> findMinHeightTrees(int n, int[][] edges) {
        List<Integer>[] graph = new List[n];
        Queue<Integer> queue = new LinkedList<Integer>();
        List<Integer> list = new ArrayList<Integer>();
        int[] degree = new int[n];
        if(n == 1) {
            list.add(0);
            return list;
        }
        if(edges == null || edges.length == 0) return list;
        for(int i = 0; i < edges.length; i++) {
            if(graph[edges[i][0]] == null) 
                graph[edges[i][0]] = new ArrayList<Integer>();
            graph[edges[i][0]].add(edges[i][1]);
            
            if(graph[edges[i][1]] == null) 
                graph[edges[i][1]] = new ArrayList<Integer>();
            graph[edges[i][1]].add(edges[i][0]);
           
            degree[edges[i][0]] ++;
            degree[edges[i][1]] ++;
        }
        for(int i = 0; i < degree.length; i++) {
            if(degree[i] == 0) return list;
            if(degree[i] == 1) queue.offer(i);
        }
        while(!queue.isEmpty()) {
            list = new ArrayList<Integer>();
            int count = queue.size();
            for(int i = 0; i < count; i++){
                int tem = queue.poll();
                list.add(tem);
                degree[tem] --;
                for(int v : graph[tem]) {
                    if(degree[v] == 0) continue;
                    if(degree[v] == 2) queue.offer(v);
                    degree[v] --;
                }
            }
        }
        return list;
    }
}