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]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if (root == null) {
            return res;
        }
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        Stack<ArrayList<Integer>> st = new Stack<ArrayList<Integer>>();
        queue.offer(root);
        int levelSize;
        while (!queue.isEmpty()) {
            levelSize = queue.size();
            ArrayList<Integer> levelres = new ArrayList<Integer>();
            for (int i = 0; i < levelSize; i++) {
                TreeNode cur  = queue.poll();
                if (cur.left != null) {
                    queue.offer(cur.left);
                }
                if (cur.right != null) {
                    queue.offer(cur.right);
                }
                levelres.add(cur.val);
            }
            st.push(new ArrayList<Integer>(levelres));
        }
        while (!st.empty()) {
            res.add(st.pop());
        }
        return res;
    }
}

同 level order traversal 一用了相同的方法，只是加了一个stack，也可以直接用 arraylist.(index, value) 方法得出结果，但是我觉得这里应该考察的是数据结构，用stack 做要好一些。

这道题目用了queue 和 stack ，注意他们增删查empty 的方法名字不一样，不要混了。