310 Minimum Height Trees

1. Question

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 containsnnodes which are labeled from0ton - 1. You will be given the numbernand a list of undirectededges(each edge is a pair of labels).
You can assume that no duplicate edges will appear inedges. Since all edges are undirected,[0, 1]is the same as[1, 0]and thus will not appear together inedges.
Example 1:
Givenn = 4,edges = [[1, 0], [1, 2], [1, 3]]
0
|
1
/ \
2 3
return[1]
Example 2:
Givenn = 6,edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]
0 1 2
\ | /
3
|
4
|
5
return[3, 4]
Note:
(1) According to the definition of tree on Wikipedia: “a tree is an undirected graph in which any two vertices are connected byexactlyone 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.

2. Implementation

(1) BFS
class Solution {
public List<Integer> findMinHeightTrees(int n, int[][] edges) {
List<Integer> res = new ArrayList<>();
if (n == 1) {
res.add(0);
return res;
}
List<Set<Integer>> adjList = new ArrayList<>();
for (int i = 0; i < n; i++) {
adjList.add(new HashSet<>());
}
for (int[] edge : edges) {
adjList.get(edge[0]).add(edge[1]);
adjList.get(edge[1]).add(edge[0]);
}
List<Integer> leaves = new ArrayList<>();
List<Integer> newLeaves = null;
for (int i = 0; i < n; i++) {
if (adjList.get(i).size() == 1) {
leaves.add(i);
}
}
while (n > 2) {
n -= leaves.size();
newLeaves = new ArrayList<>();
for (int node : leaves) {
for (int parent : adjList.get(node)) {
adjList.get(parent).remove(node);
if (adjList.get(parent).size() == 1) {
newLeaves.add(parent);
}
}
}
leaves = newLeaves;
}
return leaves;
}
}

3. Time & Space Complexity

BFS: 时间复杂度O(n), 空间复杂度O(n)