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;
}
}