1 public class Solution { 2     public List
     
       findMinHeightTrees(int n, int[][] edges) { 3         if (n == 1) { 4             return Collections.singletonList(0); 5         } 6         Map
      
       > nodes = new HashMap<>(); 7         for (int i = 0; i < n; i++) { 8             nodes.put(i, new HashSet<>()); 9         }10         for (int[] edge : edges) {11             nodes.get(edge[0]).add(edge[1]);12             nodes.get(edge[1]).add(edge[0]);13         }14         15         List
       
         leaves = new ArrayList<>();16         for (Integer i : nodes.keySet()) {17             if (nodes.get(i).size() == 1) {18                 leaves.add(i);19             }20         }21         22         while (n > 2) {23             n -= leaves.size();24             List
        
          newLeaves = new ArrayList<>();25             for (int leave : leaves) {26                 int current = nodes.get(leave).iterator().next();27                 nodes.get(current).remove(leave);28                 if (nodes.get(current).size() == 1) {29                     newLeaves.add(current);30                 }31             }32             leaves = newLeaves;33         }34         return leaves;35     }36 }
        
       
      
     

1. Check boundary case that when n == 1. There is only one node (0)

2. Set can access by iterator.