How do you solve Serialize and Deserialize BST on LeetCode?

Serialize and Deserialize BST (LeetCode #449) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding pre-order traversal is crucial for serialization.

What is the brute force solution for Serialize and Deserialize BST?

The brute force approach for Serialize and Deserialize BST is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute-force approach involves traversing the BST and converting it into a string format using a simple pre-order traversal. This method is straightforward but inefficient for larger trees. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Serialize and Deserialize BST?

Explain your thought process clearly while coding. Discuss edge cases, such as empty trees. Practice writing both serialization and deserialization together.

What are common mistakes when solving Serialize and Deserialize BST?

Forgetting to handle null nodes during serialization. Not maintaining the BST properties during deserialization.

#449

Serialize and Deserialize BST

Medium

StringTreeDepth-First SearchBreadth-First SearchDesignBinary Search TreeBinary TreeTree TraversalRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a pre-order traversal to serialize the tree while ensuring that the values are stored in a way that allows for efficient reconstruction of the BST. This method is both time-efficient and space-efficient.

⚙️

Algorithm

3 steps

1Step 1: Use a pre-order traversal to serialize the tree into a string, including markers for null nodes.
2Step 2: Split the serialized string to retrieve values during deserialization.
3Step 3: Use a recursive function to rebuild the BST based on the pre-order sequence.

solution.py25 lines

1class TreeNode:
2    def __init__(self, val=0, left=None, right=None):
3        self.val = val
4        self.left = left
5        self.right = right
6
7class Codec:
8    def serialize(self, root):
9        def preorder(node):
10            if not node:
11                return 'null'
12            return str(node.val) + ',' + preorder(node.left) + ',' + preorder(node.right)
13        return preorder(root)
14
15    def deserialize(self, data):
16        values = data.split(',')
17        def build_bst():
18            val = values.pop(0)
19            if val == 'null':
20                return None
21            node = TreeNode(int(val))
22            node.left = build_bst()
23            node.right = build_bst()
24            return node
25        return build_bst()

ℹ

Complexity note: The time complexity is O(n) because we visit each node exactly once during serialization and deserialization. The space complexity is O(n) due to the storage of the serialized string and the recursion stack.

1Understanding pre-order traversal is crucial for serialization.
2Recognizing the importance of null markers in reconstruction.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.