LC 297 - Serialize and Deserialize Binary Tree

Question

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

Links

Question here and solution here

Solution

concept

Normally we will need both inorder and preorder/postorder traversal to fully define a tree, but here we can label empty nodes as Null and this turns the tree into a Full Binary Tree and we can reconstruct the tree with just preorder traversal.

The idea is just run DFS and if we seen a null node (i.e. the current node has no child node and we still DFS down), we will append a Null in the path. We will reconstruct the tree with preorder traversal.

code

# 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):
        """Encodes a tree to a single string.
        
        :type root: TreeNode
        :rtype: str
        """
        output = []

        def dfs(node):
            if not node:
                output.append("Null")
                return
            
            output.append(str(node.val))
            dfs(node.left)
            dfs(node.right)

        dfs(root)
        return ",".join(output)
        
    def deserialize(self, data):
        """Decodes your encoded data to tree.
        
        :type data: str
        :rtype: TreeNode
        """
        self.idx = 0
        data = data.split(",")

        def dfs():
            if data[self.idx] == "Null":
                self.idx += 1
                return None

            node = TreeNode(int(data[self.idx]))
            self.idx += 1
            node.left = dfs()
            node.right = dfs()
            return node
            
        root = dfs()
        return root

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

Complexity

time: $O(n)$
space: $O(n)$