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The problem involves designing an algorithm to serialize and deserialize a binary search tree (BST). Serialization is the process of converting a BST to a string format, and deserialization is the inverse process, converting a string back to a BST.

Problem Specifications:

• Implement a class Codec with two methods:
  • serialize(root): Encodes a BST to a string.
  • deserialize(data): Decodes the string back to a BST.

class Codec:
    def serialize(self, root):
        """Encodes a BST to a single string.
        :type root: TreeNode
        :rtype: str
        """
        
    def deserialize(self, data):
        """Decodes your encoded data to tree.
        :type data: str
        :rtype: TreeNode
        """

Note:

• The input and output elements should be treated as binary search trees.
• You may assume the serialization and deserialization processes will be executed on non-null trees and the tree contains distinct values.

Clarifying Questions:

• Do we have any constraints on the size or depth of the BST?
  • No explicit constraints are given, so we assume it can fit in memory.
• Can the serialized string contain negative or very large integers?
  • Yes, it can contain any valid integer values within the normal integer range in Python.

Strategy:

The standard approach to serialize and deserialize a BST involves leveraging its properties. We’ll use Pre-order Traversal (Root-Left-Right) for serialization because it naturally fits the requirement to reconstruct the tree uniquely during deserialization due to the BST properties.

Serialize:

1. Perform a pre-order traversal of the BST.
2. Convert the tree nodes’ values to a string, separated by spaces.

Deserialize:

1. Convert the string back to a list of integers.
2. Reconstruct the BST using the list of values maintaining the BST property.

Code Implementation:

# 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 Codec:
    def serialize(self, root):
        """Encodes a BST to a single string."""
        def preorder_traverse(node):
            return [str(node.val)] + preorder_traverse(node.left) + preorder_traverse(node.right) if node else []
        
        return ' '.join(preorder_traverse(root))

    def deserialize(self, data):
        """Decodes your encoded data to tree."""
        if not data:
            return None
        
        values = list(map(int, data.split()))
        
        def build_tree(min_val, max_val):
            if values and min_val < values[0] < max_val:
                val = values.pop(0)
                node = TreeNode(val)
                node.left = build_tree(min_val, val)
                node.right = build_tree(val, max_val)
                return node
            return None
        
        return build_tree(float('-inf'), float('inf'))


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

Time Complexity:

• Serialize: O(N) where N is the number of nodes in the tree, since we visit each node once during the traversal.
• Deserialize: O(N) where N is the number of nodes in the tree, since we construct the tree by visiting each value once.

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