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 a binary tree node.
 * 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;
		}
		LinkedList<TreeNode> linkedList = new LinkedList<TreeNode>();
		linkedList.add(root);
		int cur = 1;
		int next = 0;
		ArrayList<Integer> arrayList = new ArrayList<Integer>();
		while (!linkedList.isEmpty()) {
			TreeNode first = linkedList.poll();
			cur--;
			arrayList.add(first.val);
			
			if (first.left != null) {
				linkedList.add(first.left);
				next++;
			}
			if (first.right != null) {
				linkedList.add(first.right);
				next++;
			}
			if (cur == 0) {
				cur = next;
				next = 0;
				res.add(0, arrayList);
				arrayList = new ArrayList<Integer>();
			}
		}
		return res;
    }
}