Leetcode #449: Serialize and Deserialize BST
In this guide, we solve Leetcode #449 Serialize and Deserialize BST in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Serialization is 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 search tree.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Tree, Depth-First Search, Breadth-First Search, Design, Binary Search Tree, String, Binary Tree
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
Input: root = [2,1,3]
Output: [2,1,3]
Python Solution
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Codec:
def serialize(self, root: Optional[TreeNode]) -> str:
"""Encodes a tree to a single string."""
def dfs(root: Optional[TreeNode]):
if root is None:
return
nums.append(root.val)
dfs(root.left)
dfs(root.right)
nums = []
dfs(root)
return " ".join(map(str, nums))
def deserialize(self, data: str) -> Optional[TreeNode]:
"""Decodes your encoded data to tree."""
def dfs(mi: int, mx: int) -> Optional[TreeNode]:
nonlocal i
if i == len(nums) or not mi <= nums[i] <= mx:
return None
x = nums[i]
root = TreeNode(x)
i += 1
root.left = dfs(mi, x)
root.right = dfs(x, mx)
return root
nums = list(map(int, data.split()))
i = 0
return dfs(-inf, inf)
# Your Codec object will be instantiated and called as such:
# Your Codec object will be instantiated and called as such:
# ser = Codec()
# deser = Codec()
# tree = ser.serialize(root)
# ans = deser.deserialize(tree)
# return ans
Complexity
The time complexity is O(V+E). The space complexity is O(V).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.