Leetcode: Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3   / \  9  20    /  \   15   7

return its bottom-up level order traversal as:

[  [15,7],  [9,20],  [3]]

Same as Binary Tree Level Order Traversal I, except that this time, the new level should be added to the beginning of the arraylist. 

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();        if (root == null) {            return res;        }                Queue<TreeNode> queue = new LinkedList<TreeNode>();        queue.add(root);        while (!queue.isEmpty()) {            ArrayList<Integer> level = new ArrayList<Integer>();            int size = queue.size();            for (int i = 0; i < size; i++) {                TreeNode node = queue.poll();                level.add(node.val);                if (node.left != null) {                    queue.add(node.left);                }                if (node.right != null) {                    queue.add(node.right);                }            }            res.add(0, level);        }                return res;    }}