CODE 30: Binary Tree Zigzag Level Order Traversal

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).

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

    3
   / \
  9  20
    /  \
   15   7

return its zigzag level order traversal as:

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

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>> zigzagLevelOrder(TreeNode root) {
		// Start typing your Java solution below
		// DO NOT write main() function
		if (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;
		int layer = 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>();
				if (layer % 2 == 0) {
					for (int i = layerVals.size() - 1; i >= 0; i--) {
						layerValsCpy.add(layerVals.get(i));
					}
				} else {
					layerValsCpy.addAll(layerVals);
				}
				results.add(layerValsCpy);
				layerVals.clear();
				layer++;
			}
		}
		return results;
	}