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]
]

重点就是用一个变量来记录是从左至右还是从右至左，如果从右至左，reverse一下当前level的节点就可以了。

public List<List<Integer>> zigzagLevelOrder(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;
        ArrayList<Integer> level = new ArrayList<Integer>();
        boolean fromLeftToRight = true;

        while (!queue.isEmpty()) {
            TreeNode currentNode = queue.poll();
            currentLevelCount = currentLevelCount - 1;
            //when adding from right to left order, we need to append to the first position
            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) {
                if (!fromLeftToRight) {
                    Collections.reverse(level);
                }                
                currentLevelCount = nextLevelCount;
                nextLevelCount = 0;                
                results.add(new ArrayList<Integer>(level));
                level.clear();        
                fromLeftToRight = !fromLeftToRight;
            }

        }

        return results;        
    }