102. Binary Tree Level Order Traversal

 1 class Solution {
 2     List<List<Integer>> res = new ArrayList<>();
 3     public List<List<Integer>> levelOrder(TreeNode root) {
 4         bfs(root,0);
 5         return res;
 6             
 7     }
 8     
 9     public void bfs(TreeNode root,int n) {
10         if(root == null) return;
11         if(n > res.size()) {
12             res.add(new ArrayList<Integer>());
13         }
14         res.get(n).add(root.val);
15             
16         
17         bfs(root.left,n+1);
18         bfs(root.right,n+1);
19         
20     }
21       
22 }