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]

分析：

首先笨办法，假设每个点都是root, 然后利用BFS，看那个root到最后一个leaf的高度是多少，如果比目前找到的更小，则更新，如果相同，则把那个点加到list里。

 public List<Integer> findMinHeightTrees(int n, int[][] edges) {
         List<Integer> all = new ArrayList<Integer>();
         if (n <= ) {
             all.add();
             return all;
         }

         Map<Integer, List<Integer>> map = new HashMap<>();

         for (int i = ; i < n; i++) {
             map.put(i, new ArrayList<Integer>());
         }

         for (int[] edge : edges) {
             map.get(edge[]).add(edge[]);
             map.get(edge[]).add(edge[]);
         }

         int minHeight = Integer.MAX_VALUE;
         List<Integer> temp = new ArrayList<>();
         for (int i = ; i < n; i++) {
             int height = ;
             boolean[] visited = new boolean[n];
             visited[i] = true;
             List<Integer> list = map.get(i);
             while (list.size() != ) {
                 for (Integer k : list) {
                     if (!visited[k]) {
                         visited[k] = true;
                         temp.addAll(map.get(k));
                     }
                 }
                 list = temp;
                 temp = new ArrayList<>();
                 height++;
             }
             if (height < minHeight) {
                 all.clear();
                 all.add(i);
                 minHeight = height;
             } else if (height == minHeight) {
                 all.add(i);
             }
         }
         return all;
     }

更好的方法，先构成一棵树，把数的叶子逐层的砍掉（叶子的degree为1），当这棵树只剩下2颗或者不到两颗的节点的时候，就停止。

 public class Solution {
     public List<Integer> findMinHeightTrees(int n, int[][] edges) {
         List<Integer> leaves = new ArrayList<Integer>();
         if (n <= ) {
             leaves.add();
             return leaves;
         }

         List<Set<Integer>> graph = new ArrayList<>();

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

         for (int[] edge : edges) {
             graph.get(edge[]).add(edge[]);
             graph.get(edge[]).add(edge[]);
         }

         for (int i = ; i < n; i++) {
             if (graph.get(i).size() == ) {
                 leaves.add(i);
             }
         }

         while (n > ) {
             n -= leaves.size();
             List<Integer> newLeaves = new ArrayList<>();
             for (int leave : leaves) {
                 for (int newLeaf : graph.get(leave)) {
                     graph.get(newLeaf).remove(leave);
                     if (graph.get(newLeaf).size() == ) {
                         newLeaves.add(newLeaf);
                     }
                 }
             }
             leaves = newLeaves;
         }

         return leaves;
     }
 }