103. Binary Tree Zigzag Level Order Traversal

class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> resultList = new ArrayList<>();

        Queue<TreeNode> queue = new LinkedList<>();
        if (root != null) {
            queue.offer(root);
        }

        while (!queue.isEmpty()) {
            LinkedList<Integer> numList = new LinkedList<>(); // 注意此处声明为 LinkedList 才能同时含有 Deque 与 List 接口
            boolean leftToRight = (resultList.size() & 1) == 0;
            for (int i = queue.size(); i > 0; i--) {
                TreeNode node = queue.poll();
                if (leftToRight) {
                    numList.addLast(node.val);
                } else {
                    numList.addFirst(node.val);
                }

                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
            }
            resultList.add(numList);
        }

        return resultList;
    }
}

Reference

103. Binary Tree Zigzag Level Order Traversal
剑指 Offer 32 - III. 从上到下打印二叉树 III