I am glad I was able to make a solution from my own intuition.

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Solution

from collections import deque
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def zigzagLevelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        nodes = deque([root] if root else [])
        rightToLeft = False

        result = []
        while nodes:
            level = []
            children = deque([])
            for i in range(len(nodes)):
                if rightToLeft:
                    node = nodes.pop()
                    node.right and children.appendleft(node.right)
                    node.left and children.appendleft(node.left)
                else:
                    node = nodes.popleft()
                    node.left and children.append(node.left)
                    node.right and children.append(node.right)
                level.append(node.val)
            result.append(level)
            if rightToLeft:
                nodes += children
            else:
                nodes = children + nodes
            rightToLeft = not rightToLeft
        return result

Time complexity is O(n) = n and space complexity is O(n).