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==1){
List<Integer> ans = new ArrayList<>();
ans.add(0);
return ans;
}
List<Set<Integer>> graph = new ArrayList<>();
for(int i=0; i<n; i++)
graph.add(new HashSet<>());
for(int[] e : edges){
graph.get(e[0]).add(e[1]);
graph.get(e[1]).add(e[0]);
}
Queue<Integer> queue = new LinkedList<>();
for(int i=0; i<n; i++)
if(graph.get(i).size()==1)
queue.offer(i);
while(n>2){ /* use n instead of size() */
n -= queue.size();
Queue<Integer> next = new LinkedList<>();
while(!queue.isEmpty()){
Integer u = queue.poll();
Integer v = graph.get(u).iterator().next();
graph.get(v).remove(u);
if(graph.get(v).size()==1) next.offer(v);
graph.get(u).clear();
}
queue = next;
}
return (List)queue;
}
}