[LeetCode] Binary Tree Level Order Traversal II

Total Accepted: 9875 Total
Submissions: 31987

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.

public class Solution {
public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
ArrayList<ArrayList<Integer>>   list = new ArrayList<ArrayList<Integer>>();
Queue<TreeNode>                 queue = new LinkedList<TreeNode>();
ArrayList<Integer>              levelList = new ArrayList<Integer>();
if (root == null) return list;

queue.add(root);
queue.add(null);
while (!queue.isEmpty() && queue.peek() != null) {
TreeNode top = queue.remove();
levelList.add(top.val);

if (top.left    != null) queue.add(top.left);
if (top.right   != null) queue.add(top.right);

if (queue.peek() == null) {
queue.remove();
queue.add(null);
list.add(new ArrayList<Integer>(levelList));
levelList.clear();
}
}

Collections.reverse(list);
return list;
}
}