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:

  1. class Codec:

    • Defines the Codec class, which will contain methods for serializing and deserializing a binary tree.
  2. 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.
  3. res = []

    • Initializes an empty list to hold the serialized values of the tree.
  4. def dfs(node):

    • Defines a nested helper function dfs that performs a depth-first search (DFS) to traverse the tree.
  5. if not node:

    • Checks if the current node is null. If it is, appends "N" to the result list to represent a null node.
  6. res.append(str(node.val))

    • If the node is not null, appends the node’s value to the result list.
  7. dfs(node.left) and dfs(node.right)

    • Recursively calls dfs on the left and right children of the current node.
  8. dfs(root)

    • Initiates the DFS from the root node.
  9. return ",".join(res)

    • Joins the list of serialized values into a single string separated by commas and returns it.

Deserialization Process:

  1. def deserialize(self, data: str) -> Optional[TreeNode]:

    • Defines the deserialize method, which takes a serialized string and reconstructs the binary tree.
  2. vals = data.split(",")

    • Splits the serialized string into a list of values.
  3. self.idx = 0

    • Initializes an index to track the position in the vals list during deserialization.
  4. def dfs():

    • Defines another nested function dfs to help reconstruct the tree recursively.
  5. if vals[self.idx] == "N":

    • Checks if the current value is "N". If so, increments the index and returns None, indicating a null node.
  6. node = TreeNode(int(vals[self.idx]))

    • Creates a new tree node with the value at the current index.
  7. self.idx += 1

    • Moves to the next value in the vals list.
  8. node.left = dfs() and node.right = dfs()

    • Recursively calls dfs to construct the left and right subtrees of the current node.
  9. return node

    • Returns the constructed node.
  10. 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
  1. Serialization:

    • Calling serialize(root) with root 1.
    • The function would traverse as follows:
      • 12NN3NN
    • Resulting serialized string: "1,2,N,N,3,N,N".
  2. 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:

  1. Edge Cases:

    • Consider the case where the tree is empty (i.e., the root is None).
  2. Understanding DFS:

    • Ensure you understand how depth-first search is utilized for both serialization and deserialization.
  3. 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

  1. Choose Appropriate Methods:

    • Select the method best suited to specific problem requirements to ensure efficiency and readability.
  2. Handle Edge Cases:

    • Always consider edge cases in algorithm design to avoid potential errors.
  3. Optimize Space Usage:

    • When handling large datasets, consider using space-efficient algorithms.

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