LeetCode 101: 297 Serialize and Deserialize Binary Tree

Here’s a detailed breakdown of the solution to the LeetCode problem "Serialize and Deserialize Binary Tree" (Problem 297), including explanations, examples, and insights.

LeetCode Problem: 297. Serialize and Deserialize Binary Tree

https://leetcode.com/problems/serialize-and-deserialize-binary-tree/description/

Problem Description:

Design an algorithm to serialize and deserialize a binary tree. Serialization is the process of converting a data structure into a string representation, and deserialization is the process of converting a string representation back into a data structure.

Example:

Given a binary tree, serialize it:
    1
   / \
  2   3
     / \
    4   5

The serialized format should be:
"1,2,N,N,3,4,N,N,5,N,N"

To deserialize it, you should return the binary tree.

Original Code (DFS Approach):

class Codec:

    # Encodes a tree to a single string.
    def serialize(self, root: Optional[TreeNode]) -> str:
        res = []

        def dfs(node):
            if not node:
                res.append("N")  # Use "N" for null nodes
                return
            res.append(str(node.val))  # Add the current node's value
            dfs(node.left)  # Recursively serialize the left subtree
            dfs(node.right)  # Recursively serialize the right subtree

        dfs(root)  # Start the DFS from the root
        return ",".join(res)  # Join the results into a single string

    # Decodes your encoded data to tree.
    def deserialize(self, data: str) -> Optional[TreeNode]:
        vals = data.split(",")  # Split the string into values
        self.idx = 0  # Initialize an index to track position in vals

        def dfs():
            if vals[self.idx] == "N":  # Check for null node
                self.idx += 1
                return None
            node = TreeNode(int(vals[self.idx]))  # Create a new tree node
            self.idx += 1  # Move to the next value
            node.left = dfs()  # Recursively deserialize the left subtree
            node.right = dfs()  # Recursively deserialize the right subtree
            return node  # Return the constructed node

        return dfs()  # Start the deserialization process

Line-by-Line Explanation and Example Execution:

class Codec:
- Defines the Codec class, which will contain methods for serializing and deserializing a binary tree.
def serialize(self, root: Optional[TreeNode]) -> str:
- Defines the serialize method, which takes the root of a binary tree and returns its serialized string representation.
res = []
- Initializes an empty list to hold the serialized values of the tree.
def dfs(node):
- Defines a nested helper function dfs that performs a depth-first search (DFS) to traverse the tree.
if not node:
- Checks if the current node is null. If it is, appends "N" to the result list to represent a null node.
res.append(str(node.val))
- If the node is not null, appends the node’s value to the result list.
dfs(node.left) and dfs(node.right)
- Recursively calls dfs on the left and right children of the current node.
dfs(root)
- Initiates the DFS from the root node.
return ",".join(res)
- Joins the list of serialized values into a single string separated by commas and returns it.

Deserialization Process:

def deserialize(self, data: str) -> Optional[TreeNode]:
- Defines the deserialize method, which takes a serialized string and reconstructs the binary tree.
vals = data.split(",")
- Splits the serialized string into a list of values.
self.idx = 0
- Initializes an index to track the position in the vals list during deserialization.
def dfs():
- Defines another nested function dfs to help reconstruct the tree recursively.
if vals[self.idx] == "N":
- Checks if the current value is "N". If so, increments the index and returns None, indicating a null node.
node = TreeNode(int(vals[self.idx]))
- Creates a new tree node with the value at the current index.
self.idx += 1
- Moves to the next value in the vals list.
node.left = dfs() and node.right = dfs()
- Recursively calls dfs to construct the left and right subtrees of the current node.
return node
- Returns the constructed node.
return dfs()
- Initiates the deserialization process from the start of the values list.

Example Execution:

Assuming we have the following binary tree:

    1
   / \
  2   3

Serialization:
- Calling serialize(root) with root 1.
- The function would traverse as follows:
  - 1 → 2 → N → N → 3 → N → N
- Resulting serialized string: "1,2,N,N,3,N,N".
Deserialization:
- Calling deserialize("1,2,N,N,3,N,N").
- The function reconstructs the tree:
  - Read 1, create node 1.
  - Read 2, create node 2.
  - Read N, set left of 2 to None.
  - Read N, set right of 2 to None.
  - Read 3, create node 3.
  - Read N, set left of 3 to None.
  - Read N, set right of 3 to None.
- The reconstructed tree matches the original.

Time Complexity Analysis:

Time Complexity: O(n)
- Where n is the number of nodes in the tree. Each node is processed once during serialization and deserialization.

Space Complexity Analysis:

Space Complexity: O(n)
- For the serialized string and the call stack during recursion.

Tips and Warnings:

Edge Cases:
- Consider the case where the tree is empty (i.e., the root is None).
Understanding DFS:
- Ensure you understand how depth-first search is utilized for both serialization and deserialization.
Efficiency:
- The use of lists and indices allows efficient tree reconstruction without the need for additional structures.

Summary

DFS Method: Effectively serializes and deserializes a binary tree, with a time complexity of O(n) and a space complexity of O(n).
Clear and Understandable: The code is straightforward and easy to understand, suitable for handling this type of problem.

Application Tips

Choose Appropriate Methods:
- Select the method best suited to specific problem requirements to ensure efficiency and readability.
Handle Edge Cases:
- Always consider edge cases in algorithm design to avoid potential errors.
Optimize Space Usage:
- When handling large datasets, consider using space-efficient algorithms.

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