Serialize and Deserialize Binary Tree

Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.

Design an algorithm to serialize and deserialize a binary tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure.

Clarification: The input/output format is the same as how LeetCode serializes a binary tree. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.

Example 1:

Input: root = [1,2,3,null,null,4,5]
Output: [1,2,3,null,null,4,5]

Example 2:

Input: root = []
Output: []

Constraints:

• The number of nodes in the tree is in the range [0, 104].
• -1000 <= Node.val <= 1000

SOLUTION:

# 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:
        tree = {}
        paths = [(root, 0)]
        while len(paths) > 0:
            curr, i = paths.pop()
            if curr:
                tree[i] = curr.val
                paths.append((curr.left, 2 * i + 1))
                paths.append((curr.right, 2 * i + 2))
        return ";".join(["{}={}".format(k, v) for k, v in tree.items()])

    def deserialize(self, data: str) -> Optional[TreeNode]:
        vals = data.split(";")
        tree = {}
        for v in vals:
            if len(v) > 0:
                key, value = v.split("=")
                tree[int(key)] = int(value)
        if len(tree) == 0:
            return None
        root = TreeNode()
        paths = [(root, 0)]
        while len(paths) > 0:
            curr, i = paths.pop()
            curr.val = tree[i]
            if (2 * i + 1) in tree:
                curr.left = TreeNode()
                paths.append((curr.left, 2 * i + 1))
            if (2 * i + 2) in tree:
                curr.right = TreeNode()
                paths.append((curr.right, 2 * i + 2))
        return root


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