Problem URL: Serialize and deserialize binary tree

For serialize the tree, we will traverse the tree in preorder and whenever we hit a null node, we put a # character there. Then we join each character with a comma delimeter and return the string.

For deserialize the string, first we split the string with a comma delemeter. Then we run another preorder traversal, whenever we found a # character, we put a null node there, otherwise we create a tree node with the value. Finally, we return the root node.

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Codec:
    def serialize(self, root: Optional[TreeNode]) -> str:
        res = []

        def dfs(root):
            if not root:
                res.append('#')
                return
            res.append(str(root.val))
            dfs(root.left)
            dfs(root.right)

        dfs(root)
        return ",".join(res)

    def deserialize(self, data: str) -> TreeNode:
        vals = data.split(",")
        self.i = 0

        def dfs():
            if vals[self.i] == "#":
                self.i += 1
                return None
            node = TreeNode(int(vals[self.i]))
            self.i += 1
            node.left = dfs()
            node.right = dfs()
            return node

        return dfs()

# Your Codec object will be instantiated and called as such:
# ser = Codec()
# deser = Codec()
# ans = deser.deserialize(ser.serialize(root))

Time Complexity: O(n)
Space Complexity: O(n)