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 containsnnodes which are labeled from0ton - 1. You will be given the numbernand a list of undirectededges(each edge is a pair of labels).

You can assume that no duplicate edges will appear inedges. Since all edges are undirected,[0, 1]is the same as[1, 0]and thus will not appear together inedges.

Example 1:

Givenn = 4,edges = [[1, 0], [1, 2], [1, 3]]

        0
        |
        1
       / \
      2   3

return[1]

Example 2:

Givenn = 6,edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]

     0  1  2
      \ | /
        3
        |
        4
        |
        5

return[3, 4]

Hint: at most 2 MHTs for each graph.

Analysis:

Build graph, using Map<Integer, HashSet<Integer>> or ArrayList<Set<Integer> to represent both fine.

Like peal a onion, every round delete all leaves node util there are only 2 node(or 1 node) in the graph, then return.

Code: (Space: O(n), Time: O(n))

public class Solution {
    public List<Integer> findMinHeightTrees(int n, int[][] edges) {
        if (n == 1) return Collections.singletonList(0);
        List<Integer> leaves = new ArrayList<>();
        List<Set<Integer>> adj = new ArrayList<>(n);
        for (int i = 0; i < n; ++i) adj.add(new HashSet<>());
        for (int[] edge : edges) {
            adj.get(edge[0]).add(edge[1]);
            adj.get(edge[1]).add(edge[0]);
        }
        for (int i = 0; i < n; ++i) {
            if (adj.get(i).size() == 1) leaves.add(i);
        }
        while (n > 2) {
            n -= leaves.size();
            List<Integer> newLeaves = new ArrayList<>();
            for (int i : leaves) {
                int t = adj.get(i).iterator().next();
                adj.get(t).remove(i);
                if (adj.get(t).size() == 1) newLeaves.add(t);
            }
            leaves = newLeaves;
        }
        return leaves;
    }
}