Efficient BFS solution! 
IntuitionThe zigzag level order traversal is a variation of the level order traversal (BFS). The key difference is alternating the order of nodes at each level:
# Intuition\nThe zigzag level order traversal is a variation of the level order traversal (BFS). The key difference is alternating the order of nodes at each level: left-to-right for one level and right-to-left for the next. We can achieve this by traversing each level normally and reversing the values on alternating levels.\n\n# Approach\n1. Use a `deque` to perform a BFS traversal of the tree.\n2. Maintain a boolean flag `zigzag` to track whether the current level should be reversed.\n3. For each level:\n   - Traverse all the nodes in the current level, appending their values to a temporary list.\n   - If `zigzag` is `True`, reverse the list before appending it to the final result.\n   - Add the children of the nodes to the queue for the next level.\n4. Toggle the `zigzag` flag after processing each level.\n5. Return the final result containing all levels in the desired zigzag order.\n\n# Complexity\n- Time complexity:\nO(n), where \\(n\\) is the number of nodes in the tree.\n\n- Space complexity:\nThe queue can hold up to \\(O(n)\\) nodes when the tree is completely balanced at its last level.\n\n\n# Code\n```python3 []\n# Definition for a binary tree node.\n# class TreeNode:\n#     def __init__(self, val=0, left=None, right=None):\n#         self.val = val\n#         self.left = left\n#         self.right = right\nclass Solution:\n    def zigzagLevelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:\n        if root is None: return []\n\n        queue = deque()\n        queue.append(root)\n        ans = []\n        zigzag = False\n\n        while queue:\n            level = []\n            n = len(queue)\n\n            for i in range(n):\n                node = queue.popleft()\n                level.append(node.val)\n\n                if node.left: queue.append(node.left)\n                if node.right: queue.append(node.right)\n            \n            if zigzag:\n                level.reverse()\n            \n            ans.append(level)\n            zigzag = not zigzag\n\n        return ans\n\n```
['Python3']