102. Binary Tree Level Order Traversal
身体不好
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 }