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,null,null,15,7],
    3
   / \
  9  20
    /  \
   15   7
return its zigzag level order traversal as:
[
  [3],
  [20,9],
  [15,7]
]

难度：medium

题目：给定二叉树，返回其之字型层次遍历结点值。

思路：层次遍历

Runtime: 1 ms, faster than 90.32% of Java online submissions for Binary Tree Zigzag Level Order Traversal.
Memory Usage: 37.1 MB, less than 100.00% of Java online submissions for Binary Tree Zigzag Level Order Traversal.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        if (null == root) {
            return result;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        int currentLevelCount = 1, nextLevelCount = 0;
        boolean direction = true;
        while (!queue.isEmpty()) {
            List<Integer> curLevelElems = new ArrayList<>(currentLevelCount);
            for (int i = 0; i < currentLevelCount; i++) {
                TreeNode node = queue.poll();
                if (node.left != null) {
                    queue.add(node.left);
                    nextLevelCount++;
                }
                if (node.right != null) {
                    queue.add(node.right);
                    nextLevelCount++;
                }
                if (direction) {
                    curLevelElems.add(node.val);
                } else {
                    curLevelElems.add(0, node.val);
                }
            }
            result.add(curLevelElems);
            currentLevelCount = nextLevelCount;
            nextLevelCount = 0;
            direction = !direction;
        }
        
        return result;
    }
}