LeetCode Graph：M310. Minimum Height Trees

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]

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.

Credits: Special thanks to @dietpepsi for adding this problem and creating all test cases.

//一般给的是边，不方便用BFS，更适合拓扑 //利用拓扑的性质 public class Solution { int[][] es; public List<Integer> findMinHeightTrees(int n, int[][] edges) { es = edges; List<Integer> res = new ArrayList<Integer>(); if(n==0) return res; if(n==1){ res.add(0); return res; } //n>1 int m = es.length;//边的个数 int[] nums = new int[n];//保存连接到点i上的点的个数 //计算连接到点i上的点的个数 for(int i=0; i<m; i++){ nums[es[i][0]]++; nums[es[i][1]]++; } int count = 0; while(true){ //找到所有的叶子,度=1 Queue<Integer> q = new LinkedList<Integer>(); for(int i=0; i<n; i++){ if(nums[i]==1){ q.add(i); count++; nums[i]=-1; }else if(nums[i]==0){ res.add(i); return res; } } //最后 if(count==n){ while(!q.isEmpty()) res.add(q.poll()); return res; } //删除所有的叶子 while(!q.isEmpty()){ int t = q.poll(); for(int i=0; i<m; i++){ int a = es[i][0]; int b = es[i][1]; if((a==t || b==t)){ int tmp; if(a==t) tmp = b; else tmp = a ; if(nums[tmp]!=-1){ nums[tmp]--; break; } } } } } } }