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:

public class Solution {
public List<Integer> findMinHeightTrees(int n, int[][] edges) {
List<Integer> res = new ArrayList<Integer>();
if(n <= 1 || edges == null || edges.length == 0 || edges[0].length == 0){
res.add(0);
return res;
}
Map<Integer, Set<Integer>> graph = new HashMap<Integer, Set<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(int i = 0; i<n; i++){
if(graph.get(i).size() == 1){
que.add(i);
}
}

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

for(int cur : que){
for(int neighbour : graph.get(cur)){
graph.get(cur).remove(neighbour);
graph.get(neighbour).remove(cur);
if(graph.get(neighbour).size() == 1){
newQue.add(neighbour);
}
}
}

que = newQue;
}

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