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Serialize and Deserialize BST

Design an algorithm to serialize and deserialize a binary search tree with efficient traversal and state tracking.

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Practice Focus

Medium · Binary-tree traversal and state tracking

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Answer-first summary

Design an algorithm to serialize and deserialize a binary search tree with efficient traversal and state tracking.

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The problem requires designing an algorithm to serialize and deserialize a binary search tree (BST). Serialization converts a BST into a string representation, which can then be deserialized back into the original tree structure. Focus on binary-tree traversal methods and state tracking during the process to ensure correctness and efficiency.

Problem Statement

Given a binary search tree (BST), your task is to design an algorithm for serializing and deserializing the tree. Serialization refers to converting the BST into a compact string that can be easily stored or transmitted, while deserialization refers to reconstructing the tree from this string.

Your algorithm must ensure that the binary search tree can be correctly serialized and deserialized, maintaining the tree structure with no loss of information. The encoded string should be as compact as possible, and you are allowed to use any serialization or deserialization approach, as long as the process is efficient.

Examples

Example 1

Input: root = [2,1,3]

Output: [2,1,3]

Example details omitted.

Example 2

Input: root = []

Output: []

Example details omitted.

Constraints

  • The number of nodes in the tree is in the range [0, 104].
  • 0 <= Node.val <= 104
  • The input tree is guaranteed to be a binary search tree.

Solution Approach

Pre-order Traversal with String Construction

A common approach is to perform a pre-order traversal (root, left, right) of the tree while appending node values to a string. You can use markers (e.g., 'null') to represent null nodes for completeness. During deserialization, the string is split and the tree is reconstructed by recursively inserting nodes into the tree according to the pre-order traversal.

Breadth-First Search (BFS) with Queue

Another approach uses a breadth-first search to traverse the tree level by level. A queue can be used to manage the nodes during both serialization and deserialization. In serialization, each node is enqueued and added to a string, and in deserialization, the string is used to reconstruct the tree by processing nodes in the order they were serialized.

Space Optimized with Iterative DFS

For more memory-efficient solutions, you can use an iterative depth-first search (DFS) with a stack to simulate the recursion. This approach minimizes the space complexity by avoiding the overhead of recursive calls and is particularly useful when dealing with large trees.

Complexity Analysis

Metric Value
Time Depends on the final approach
Space Depends on the final approach

The time and space complexity depend on the traversal approach used. For pre-order or BFS, both time and space complexity are O(N), where N is the number of nodes in the tree. In the space-optimized approach, the space complexity can be reduced to O(H), where H is the height of the tree, but the time complexity remains O(N).

What Interviewers Usually Probe

  • Look for the candidate's understanding of tree traversal methods and how they relate to serialization.
  • Evaluate the ability to choose the most space-efficient approach based on the tree's size and structure.
  • Assess the candidate's attention to detail in handling edge cases, such as empty trees or trees with a large number of nodes.

Common Pitfalls or Variants

Common pitfalls

  • Over-complicating the algorithm by not considering simple traversal techniques like pre-order or BFS.
  • Failing to account for null nodes during serialization, leading to incorrect deserialization.
  • Not optimizing for space in scenarios with large trees, resulting in excessive memory usage.

Follow-up variants

  • Using a post-order or in-order traversal for serialization instead of pre-order.
  • Implementing the solution iteratively without recursion.
  • Optimizing the solution for trees with high depth and low width.

FAQ

What is the best traversal method for serializing a BST?

Pre-order traversal is commonly used for serializing a BST because it preserves the root and subtree relationships, making deserialization easier.

How can I optimize the space complexity of this problem?

You can optimize space by using iterative depth-first search (DFS) to avoid recursive stack overhead, reducing the space complexity.

Can the algorithm handle empty trees?

Yes, the algorithm should be designed to handle empty trees, typically by encoding them as an empty string or 'null' values.

What is the expected time complexity for this problem?

The time complexity is O(N) for both serialization and deserialization, where N is the number of nodes in the tree.

What happens if I use in-order traversal for serialization?

In-order traversal doesn't maintain the structure of a BST for serialization, so it's not suitable for this problem where you need to reconstruct the tree precisely.

terminal

Solution

Solution 1

#### Python3

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# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


class Codec:
    def serialize(self, root: Optional[TreeNode]) -> str:
        """Encodes a tree to a single string."""

        def dfs(root: Optional[TreeNode]):
            if root is None:
                return
            nums.append(root.val)
            dfs(root.left)
            dfs(root.right)

        nums = []
        dfs(root)
        return " ".join(map(str, nums))

    def deserialize(self, data: str) -> Optional[TreeNode]:
        """Decodes your encoded data to tree."""

        def dfs(mi: int, mx: int) -> Optional[TreeNode]:
            nonlocal i
            if i == len(nums) or not mi <= nums[i] <= mx:
                return None
            x = nums[i]
            root = TreeNode(x)
            i += 1
            root.left = dfs(mi, x)
            root.right = dfs(x, mx)
            return root

        nums = list(map(int, data.split()))
        i = 0
        return dfs(-inf, inf)


# Your Codec object will be instantiated and called as such:
# Your Codec object will be instantiated and called as such:
# ser = Codec()
# deser = Codec()
# tree = ser.serialize(root)
# ans = deser.deserialize(tree)
# return ans
Serialize and Deserialize BST Solution: Binary-tree traversal and state track… | LeetCode #449 Medium