标签：
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:
1 public class Solution { 2 public List<Integer> findMinHeightTrees(int n, int[][] edges) { 3 List<Integer> res = new ArrayList<Integer>(); 4 if(n <= 1 || edges == null || edges.length == 0 || edges[0].length == 0){ 5 res.add(0); 6 return res; 7 } 8 Map<Integer, Set<Integer>> graph = new HashMap<Integer, Set<Integer>>(); 9 for(int i = 0; i<n; i++){ 10 graph.put(i, new HashSet<Integer>()); 11 } 12 for(int [] edge : edges){ 13 graph.get(edge[0]).add(edge[1]); 14 graph.get(edge[1]).add(edge[0]); 15 } 16 17 LinkedList<Integer> que = new LinkedList<Integer>(); 18 for(int i = 0; i<n; i++){ 19 if(graph.get(i).size() == 1){ 20 que.add(i); 21 } 22 } 23 24 while(n > 2){ 25 n -= que.size(); 26 LinkedList<Integer> newQue = new LinkedList<Integer>(); 27 28 for(int cur : que){ 29 for(int neighbour : graph.get(cur)){ 30 graph.get(cur).remove(neighbour); 31 graph.get(neighbour).remove(cur); 32 if(graph.get(neighbour).size() == 1){ 33 newQue.add(neighbour); 34 } 35 } 36 } 37 38 que = newQue; 39 } 40 41 res.addAll(que); 42 return res; 43 } 44 }
标签：
原文地址：http://www.cnblogs.com/Dylan-Java-NYC/p/5301334.html
