Leetcode 294: Serialize and Deserialize - DSA Interview Preparation

Let's dive into the solution for Leetcode's problem 297, "Serialize and Deserialize Binary Tree," covering all aspects from understanding the question to efficient code explanations.

Understanding the Problem Statement

The problem asks us to design an algorithm to serialize and deserialize a binary tree. In simpler terms:

Serialize: Convert a binary tree to a string representation.
Deserialize: Convert the string back to its original binary tree structure.

The problem is essentially about converting the binary tree into a string (serialization) and then reconstructing it from that string (deserialization). The objective is to ensure that you can perfectly reconstruct the original tree from the serialized data.

Brute Force Approach

A common brute force approach for this problem might be to use an in-order or pre-order traversal of the tree to record node values. For instance:

Traverse each node of the binary tree, storing the values in an array or string.
Include special symbols to represent null values, which can help distinguish between different shapes of the tree.

However, this method has its limitations:

It doesn't always capture the full tree structure.
It may require additional work to keep track of missing nodes.
Handling a binary tree's hierarchical nature in a linear format can be cumbersome.

Hint to Solve the Problem Efficiently

To solve this problem efficiently, think in terms of Breadth-First Search (BFS), which will allow you to traverse the tree level by level. This approach ensures that all nodes are visited in the correct order, and you can easily capture null children with a placeholder (like "N" in our solution). The use of a queue data structure will help manage the nodes during traversal.

Efficient Solution

The provided code utilizes a BFS approach to serialize and deserialize the binary tree. Let's break down each part of the solution.

Serialization

The serialize function converts the binary tree into a single string.

BFS Traversal: Starting from the root, we traverse level by level.
Queue Management: A queue is used to manage nodes at each level.
Null Nodes: If a node is None, we add "N" to represent it. Otherwise, we add the node's value and push its children to the queue.
String Construction: The values are joined with commas to form the final string representation.

Code for Serialization:

pythonCopy codedef serialize(self, root):
    if not root:
        return ""
    queue = [root]
    result = []
    while queue:
        node = queue.pop(0)
        if node:
            result.append(str(node.val))
            queue.append(node.left)
            queue.append(node.right)
        else:
            result.append("N")
    return ",".join(result)

In this approach:

Nodes are stored sequentially in a string.
N is used to represent None nodes, ensuring that we capture the exact structure.

Deserialization

The deserialize function reconstructs the binary tree from the string representation.

Data Splitting: The string is split by commas to get individual node values.
Queue Management: We use a queue to manage the tree construction.
Node Reconstruction: Starting from the root, we reconstruct the left and right children for each node using BFS.

Code for Deserialization:

pythonCopy codedef deserialize(self, data):
    if not data:
        return None
    values = data.split(",")
    root = TreeNode(int(values[0]))
    queue = [root]
    i = 1
    while queue:
        node = queue.pop(0)
        if values[i] != "N":
            node.left = TreeNode(int(values[i]))
            queue.append(node.left)
        i += 1
        if values[i] != "N":
            node.right = TreeNode(int(values[i]))
            queue.append(node.right)
        i += 1
    return root

Here:

The string is split to reconstruct individual nodes.
The BFS approach is used again to construct the tree level by level, ensuring that each node is appropriately linked.

Time and Space Complexity

Time Complexity: Both serialization and deserialization have a time complexity of O(N), where N is the number of nodes in the binary tree. This is because each node is visited exactly once during both processes.
Space Complexity: The space complexity is also O(N), where N is the number of nodes in the binary tree. This is due to:
- The space required to store the serialized output.
- The queue used in both serialization and deserialization to manage nodes.

Conclusion

This BFS-based approach is efficient and ensures that the serialized data retains all structural information necessary for perfect reconstruction. It uses a queue to systematically traverse and manage the nodes, making the solution straightforward and effective.

If you have any questions or need further clarification, feel free to reach out!

Serialize and Deserialize Binary Tree (Leetcode #297)

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