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:

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.

Solution

public class Solution {
    public List<Integer> findMinHeightTrees(int n, int[][] edges) {
        if(n == 0 || edges == null) return new ArrayList<Integer>();
        if(n == 1) return Arrays.asList( new Integer[]{0} );

        Map<Integer, List<Integer>> edge = new HashMap<Integer, List<Integer>>();
        for(int i = 0; i < edges.length; i++) {
            int from = edges[i][0];
            int to = edges[i][1];

            if(edge.get(from) == null) edge.put(from, new ArrayList<Integer>());
            if(edge.get(to) == null) edge.put(to, new ArrayList<Integer>());
            edge.get(from).add(to);
            edge.get(to).add(from);
        }


        List<Integer> path1 = dfs(edge, 0);
        List<Integer> path2 = dfs(edge, path1.get(path1.size()-1));

        // get the middle one
        List<Integer> ans = new ArrayList<Integer>();
        if(path2.size() % 2 == 0) {
            ans.add( path2.get(path2.size()/2)  );
            ans.add( path2.get(path2.size()/2-1));
        } else {
            ans.add( path2.get(path2.size()/2)  );
        }
        return ans;
    }

    public List<Integer> dfs(Map<Integer, List<Integer>> edge, int from) {
        Set<Integer> visited = new HashSet<Integer>();

        List<Integer> longest = new ArrayList<Integer>();
        List<Integer> path = new ArrayList<Integer>();
        path.add(from);
        visited.add(from);
        while(path.size() != 0) {
            int cur = path.get(path.size()-1);

            boolean block = true;
            // if no way to go, pop out
            for(Integer to: edge.get(cur)) {
                if(visited.add(to)) {
                    path.add(to);
                    block = false;
                    break;
                }
            }

            if(block) {
                if(path.size() > longest.size()) longest = new ArrayList<Integer>(path);
                path.remove(path.size()-1);
            }
        }

        return longest;
    }

}