LeetCode Minimum Height Trees

原题链接在这里：https://leetcode.com/problems/minimum-height-trees/

题目：

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]

题解：

与Course Schedule, Course Schedule II类似。

用BFS based topological sort. 从叶子节点开始放入queue中，最后剩下的一个或者两个就是最中心的点.

这里练习undirected graph的topological sort. 用Map<Integer, Set<Integer>>来表示graph, 一条edge, 两头node都需要加graph中.

Time Complexity: O(n+e). Space: O(n+e).

AC Java:

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

         if(n < 1 || edges == null || edges.length == 0){
             return res;
         }

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

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

         LinkedList<Integer> que = new LinkedList<Integer>();
         for(Map.Entry<Integer, HashSet<Integer>> entry : graph.entrySet()){
             if(entry.getValue().size() == 1){
                 que.add(entry.getKey());
             }
         }

         while(n > 2){
             n -= que.size();
             LinkedList<Integer> temp = new LinkedList<Integer>();

             while(!que.isEmpty()){
                 int cur = que.poll();
                 for(int neigh : graph.get(cur)){
                     graph.get(cur).remove(neigh);
                     graph.get(neigh).remove(cur);

                     if(graph.get(neigh).size() == 1){
                         temp.add(neigh);
                     }
                 }
             }

             que = temp;
         }

         res.addAll(que);
         return res;
     }
 }