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 = 4edges = [[1, 0], [1, 2], [1, 3]]

        0        |        1       / \      2   3

return [1]

Example 2:

Given n = 6edges = [[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;                      }                  }                                          }           }                 }  }}