LeetCode Graph:M310. Minimum Height Trees
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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; } } } } } }}
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