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 <= 1) {
            all.add(0);
            return all;
        }

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

        for (int i = 0; i < n; i++) {
            map.put(i, new ArrayList<Integer>());
        }

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

        int minHeight = Integer.MAX_VALUE;
        List<Integer> temp = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            int height = 0;
            boolean[] visited = new boolean[n];
            visited[i] = true;
            List<Integer> list = map.get(i);
            while (list.size() != 0) {
                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 <= 1) {
            leaves.add(0);
            return leaves;
        }

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

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

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

        for (int i = 0; i < n; i++) {
            if (graph.get(i).size() == 1) {
                leaves.add(i);
            }
        }

        while (n > 2) {
            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() == 1) {
                        newLeaves.add(newLeaf);
                    }
                }
            }
            leaves = newLeaves;
        }

        return leaves;
    }
}