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]
一道图的题目,我们可以每次都删除叶子节点,维护一个入度为1的队列,删除叶子节点的同时,将新的入度为0的及诶单更新到队列中,直到最后剩下一个或两个节点。代码如下:
public class Solution { public List<Integer> findMinHeightTrees(int n, int[][] edges) { List<Integer>[] graph = new List[n]; Queue<Integer> queue = new LinkedList<Integer>(); List<Integer> list = new ArrayList<Integer>(); int[] degree = new int[n]; if(n == 1) { list.add(0); return list; } if(edges == null || edges.length == 0) return list; for(int i = 0; i < edges.length; i++) { if(graph[edges[i][0]] == null) graph[edges[i][0]] = new ArrayList<Integer>(); graph[edges[i][0]].add(edges[i][1]); if(graph[edges[i][1]] == null) graph[edges[i][1]] = new ArrayList<Integer>(); graph[edges[i][1]].add(edges[i][0]); degree[edges[i][0]] ++; degree[edges[i][1]] ++; } for(int i = 0; i < degree.length; i++) { if(degree[i] == 0) return list; if(degree[i] == 1) queue.offer(i); } while(!queue.isEmpty()) { list = new ArrayList<Integer>(); int count = queue.size(); for(int i = 0; i < count; i++){ int tem = queue.poll(); list.add(tem); degree[tem] --; for(int v : graph[tem]) { if(degree[v] == 0) continue; if(degree[v] == 2) queue.offer(v); degree[v] --; } } } return list; } }