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]

Hint:

1. How many MHTs can a graph have at most?

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 @peisi for adding this problem and creating all test cases.

public class Solution {
    public List<Integer> findMinHeightTrees(int n, int[][] edges) {
        List<Integer> res = new ArrayList<>();
        if (n == 0) {
            return res;
        }
        if (n == 1) {
            res.add(0);
            return res;
        }
        List<Set<Integer>> adjacent = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adjacent.add(new HashSet<Integer>());
        }
        for (int[] edge : edges) {
            int from = edge[0];
            int to = edge[1];
            adjacent.get(from).add(to);
            adjacent.get(to).add(from);
        }
        LinkedList<Integer> leaves = new LinkedList<>();
        for (int i = 0; i < n; i++) {
            if (adjacent.get(i).size() == 1) {
                leaves.add(i);
            }
        }
        int numOfLeaves = n;
        while (numOfLeaves > 2) {
            numOfLeaves -= leaves.size();
            int count = leaves.size();
            while (count-- > 0) {
                int from = leaves.poll();
                int to = adjacent.get(from).iterator().next();
                adjacent.get(to).remove(from);
                if (adjacent.get(to).size() == 1) {
                    leaves.add(to);
                }
            }
        }
        return leaves;
    }
}