How do you solve Serialize and Deserialize Binary Tree on LeetCode?

Serialize and Deserialize Binary Tree (LeetCode #297) 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 tree traversal is crucial for serialization and deserialization.

What is the brute force solution for Serialize and Deserialize Binary Tree?

The brute force approach for Serialize and Deserialize Binary Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves traversing the tree and converting it into a string format by using a simple pre-order traversal. This method is straightforward but inefficient because it repeatedly constructs the string and may involve unnecessary operations. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Serialize and Deserialize Binary Tree?

Explain your thought process clearly while coding. Discuss edge cases, such as empty trees, before diving into implementation. Practice writing both serialization and deserialization together to reinforce understanding.

What are common mistakes when solving Serialize and Deserialize Binary Tree?

Failing to handle null nodes correctly during serialization. Not maintaining the order of nodes during deserialization, leading to incorrect tree structure.

#297

Serialize and Deserialize Binary Tree

Hard

StringTreeDepth-First SearchBreadth-First SearchDesignBinary TreeTree TraversalRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses a pre-order traversal to serialize the tree into a string and utilizes an iterator to deserialize it back into the tree structure. This method is efficient and straightforward, leveraging the properties of tree traversal.

⚙️

Algorithm

3 steps

1Step 1: Serialize the tree using pre-order traversal, appending values to a string with commas separating them.
2Step 2: For null nodes, append 'null' to indicate absence.
3Step 3: Deserialize by splitting the string and using an iterator to reconstruct the tree recursively.

solution.py25 lines

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

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 recursive call stack.

1Understanding tree traversal is crucial for serialization and deserialization.
2Using a queue or iterator can simplify the reconstruction process during deserialization.

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