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]
]

Trick is when adding the current level to the results, always add it to the first element in the list.

public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> results = new ArrayList<List<Integer>>();

        if (root==null) {
            return results;
        }

        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);

        int currentLevelCount = 1;
        int nextLevelCount = 0;
        List<Integer> level = new ArrayList<Integer>();

        while (!queue.isEmpty()) {
            TreeNode currentNode = queue.poll();
            currentLevelCount = currentLevelCount - 1;
            level.add(currentNode.val);
            if (currentNode.left!=null) {
                queue.add(currentNode.left);
                nextLevelCount++;
            }

            if (currentNode.right!=null) {
                queue.add(currentNode.right);
                nextLevelCount++;
            }

            if (currentLevelCount==0) {
                currentLevelCount = nextLevelCount;
                nextLevelCount = 0;    
                //the trick is here, always add the current level to the first position.
                results.add(0, new ArrayList<Integer>(level));
                level.clear();
            }    
        }

        return results;        
    }