leetcode 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]

Show Hint 

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.

Credits:

Special thanks to @dietpepsi for adding this problem and creating all test cases.

提示说一棵树中最多有几颗MHT，答案是2，这个是可以证明的，但是博主并不会- -依照这个思路我们可以每次删去度数为1的叶结点，直到树中只剩1-2个结点即为所求，代码参考的https://discuss.leetcode.com/topic/48519/java-bfs-like-ac-solution-using-user-defined-graph-node-class，我觉得比我自己写的要清晰：

public List<Integer> findMinHeightTrees(int n, int[][] edges) {        Map<Integer, Node> graph = new HashMap<>();        for (int i = 0; i < n; i++) {            graph.put(i, new Node(i));        }        for (int[] edge : edges) {            graph.get(edge[0]).neighbors.add(graph.get(edge[1]));            graph.get(edge[1]).neighbors.add(graph.get(edge[0]));            graph.get(edge[0]).degree++;            graph.get(edge[1]).degree++;        }        Queue<Node> queue = new LinkedList<>();        for (int index : graph.keySet()) {            if (graph.get(index).degree == 1) {                queue.offer(graph.get(index));            }        }        while (n > 2) {            int size = queue.size();            for (int i = 0; i < size; i++) {                Node leaf = queue.poll();                Node neighbor = leaf.neighbors.iterator().next();                neighbor.neighbors.remove(leaf);//remove leaf from its neighbor's adj list                graph.remove(leaf.label);//remove leaf self                n--;                if (--neighbor.degree == 1) {                    queue.offer(neighbor);                }            }        }        return new ArrayList<>(graph.keySet());    }}class Node {    int label;    int degree;    Set<Node> neighbors;    public Node(int index) {        label = index;        neighbors = new HashSet<>();    }