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。