<p>Given the <code>root</code> of a binary tree, return <em>the zigzag level order traversal of its nodes&#39; values</em>. (i.e., from left to right, then right to left for the next level and alternate between).</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/tree1.jpg" style="width: 277px; height: 302px;" />
<pre>
<strong>Input:</strong> root = [3,9,20,null,null,15,7]
<strong>Output:</strong> [[3],[20,9],[15,7]]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = [1]
<strong>Output:</strong> [[1]]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> root = []
<strong>Output:</strong> []
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 2000]</code>.</li>
	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
</ul>

class Solution:
    def zigzagLevelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if root is None:
            return
        result, zigzagDirection = [], 1
        q = [root]
        while q:
            level, queueLength = [], len(q)
            for i in range(queueLength):
                node = q.pop(0)
                level.append(node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            result.append(level[::zigzagDirection])
            zigzagDirection *= -1
        return result
16500