Introduction

In this blog post, we will explore how to solve the LeetCode problem "297" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

• Serialization: Perform a Breadth-First Search (BFS) traversal of the binary tree. Represent null nodes with a special character (e.g., 'N'). Convert the tree structure to a string by concatenating the values of nodes encountered during BFS, separated by a delimiter (e.g., ',').
  • Deserialization: Split the serialized string into a list of node values based on the delimiter. Reconstruct the tree level by level using the BFS approach. For each node, create its left and right children based on the corresponding values in the list. If a value is 'N', treat it as a null node.
  • Handling Empty Trees: If the input tree is empty, serialize it to an empty string and deserialize an empty string to an empty tree.

• Time and Space Complexity: O(N), where N is the number of nodes in the binary tree. This is because both serialization and deserialization involve visiting each node once.

Code

    from collections import deque

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class Codec:

    def serialize(self, root):
        """Encodes a tree to a single string.

        :type root: TreeNode
        :rtype: str
        """
        if not root:
            return ""

        queue = deque([root])
        result = []

        while queue:
            node = queue.popleft()

            if node:
                result.append(str(node.val))
                queue.append(node.left)
                queue.append(node.right)
            else:
                result.append('N')

        return ','.join(result)

    def deserialize(self, data):
        """Decodes your encoded data to tree.

        :type data: str
        :rtype: TreeNode
        """
        if not data:
            return None

        nodes = data.split(',')
        root = TreeNode(int(nodes[0]))
        queue = deque([root])
        i = 1

        while queue:
            node = queue.popleft()

            if nodes[i] != 'N':
                node.left = TreeNode(int(nodes[i]))
                queue.append(node.left)
            i += 1

            if nodes[i] != 'N':
                node.right = TreeNode(int(nodes[i]))
                queue.append(node.right)
            i += 1

        return root