[Leetcode] 102. Binary Tree Level Order Traversal (JAVA)

Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level).

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: [[3],[9,20],[15,7]]

Example 2:

Input: root = [1]
Output: [[1]]

Example 3:

Input: root = []
Output: []

Constraints:

The number of nodes in the tree is in the range [0, 2000].
-1000 <= Node.val <= 1000

My Code

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private List<List<Integer>> ret;
    public List<List<Integer>> levelOrder(TreeNode root) {
        ret = new ArrayList<>();
        helper(root, 0);
        return ret;
    }
    public void helper(TreeNode node, int depth) {
        if(node==null) return;
        if(ret.size()<=depth) ret.add(new ArrayList<Integer>());
        ret.get(depth).add(node.val);
        helper(node.left, depth+1);
        helper(node.right, depth+1);
        return;
    }
}