CODE 26: 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.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1  / \ 2   3    /   4    \     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {// Start typing your Java solution below// DO NOT write main() functionif (null == root) {return new ArrayList<ArrayList<Integer>>();}ArrayList<ArrayList<Integer>> results = new ArrayList<ArrayList<Integer>>();Queue<TreeNode> queue = new LinkedList<TreeNode>();ArrayList<Integer> layerVals = new ArrayList<Integer>();queue.offer(root);int layerNumber = 1;while (!queue.isEmpty()) {TreeNode node = queue.poll();layerVals.add(node.val);layerNumber--;if (null != node.left) {queue.offer(node.left);}if (null != node.right) {queue.offer(node.right);}if (layerNumber == 0) {layerNumber = queue.size();ArrayList<Integer> layerValsCpy = new ArrayList<Integer>();layerValsCpy.addAll(layerVals);results.add(0, layerValsCpy);layerVals.clear();}}return results;}