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Serialize and Deserialize BST

#1726 Most AskedMedium

Serialize and Deserialize BST #1726 Most Asked

Medium

Google

5 views

Topics:

TreesRecursion

Serialization is converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.

Design an algorithm to serialize and deserialize a binary search tree. There is no restriction on how your serialization/deserialization algorithm should work. You need to ensure that a binary search tree can be serialized to a string, and this string can be deserialized to the original tree structure.

The encoded string should be as compact as possible.

Example 1:

Input: root = [2,1,3]
Output: [2,1,3]

Example 2:

Input: root = []
Output: []

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.

Try coding it in LeetCode

Serialize and Deserialize BST

#1726 Most AskedMedium

Serialize and Deserialize BST #1726 Most Asked

Medium

Google

5 views

Topics:

TreesRecursion

The encoded string should be as compact as possible.

Example 1:

Input: root = [2,1,3]
Output: [2,1,3]

Example 2:

Input: root = []
Output: []

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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What data type are the node values? Are negative values allowed?
Can the BST be empty (null root)?
Does the serialization need to be space-optimal, or is readability/simplicity prioritized?
If there are multiple valid BSTs that can be constructed from the serialized string, is any valid BST acceptable upon deserialization?
Are there any constraints on the depth or number of nodes in the BST?

Brute Force Solution

Approach

The brute force method for serializing a binary search tree is to explore all possible ways to represent the tree's structure and values. We essentially try every combination until we find a valid representation that can be later reconstructed into the original tree. This exhaustive search guarantees finding a solution, though it may not be the most efficient.

Here's how the algorithm would work step-by-step:

Start by exploring representing the tree node by node.
Consider different ways to represent the value of each node and whether the node has children.
Try all the possible orders to list out each of the node values.
For each order, check if it is a valid ordering to reconstruct the tree.
Consider different ways to represent null or non-existent nodes.
Keep track of all the valid representations.
Among all the valid representations, pick one as the final serialized string.
To deserialize, we'd then attempt to rebuild the BST from each possible structure represented by the string, making sure it is actually valid.

Code Implementation

class TreeNode:
    def __init__(self, value):
        self.value = value
        self.left = None
        self.right = None

class Solution:
    def serialize(self, root):
        # Brute force serialization - generate all possibilities
        if not root:
            return ""

        nodes = []
        def traverse(node):
            if not node:
                nodes.append(None)
                return
            nodes.append(node.value)
            traverse(node.left)
            traverse(node.right)

        traverse(root)

        # Representation of all possible node orderings as string
        return ','.join(map(str, [node if node is None else node for node in nodes]))

    def deserialize(self, data):
        # Brute force deserialization - try all tree structures
        if not data:
            return None

        values = data.split(',')
        self.index = 0

        def build_tree():
            # Reconstruct the tree using recursive calls
            if self.index >= len(values) or values[self.index] == 'None':
                self.index += 1
                return None

            current_value = int(values[self.index])
            self.index += 1
            node = TreeNode(current_value)

            # Recursively build left and right subtrees
            node.left = build_tree()

            # Guaranteeing correct BST properties is crucial
            node.right = build_tree()
            return node

        return build_tree()

Big(O) Analysis

Time Complexity

O(∞) – The proposed brute force method explores all possible tree structures and node value combinations. Representing 'n' nodes in different orders involves checking n! (n factorial) permutations. For each permutation, verifying if it represents a valid BST requires significant computation, potentially involving tree construction and validation. Since n! grows faster than any polynomial function, and the validation step itself doesn't have a clear polynomial bound, the algorithm can take an astronomically long time even for modest 'n', making it practically infinite from a computational perspective and certainly not polynomial or even exponential in the typical sense. The method involves generating and testing all possible arrangements, leading to extremely high complexity.

Space Complexity

O(N!) – The brute force approach explores all possible ways to represent the tree, which involves generating all permutations of node values and structures. The number of possible permutations of N nodes can grow up to N! In addition, keeping track of all valid representations, as the algorithm dictates, requires storing potentially all N! valid strings/data structures in memory. Therefore, the space complexity is dominated by the need to store these numerous representations, resulting in O(N!) auxiliary space, where N is the number of nodes in the BST.

Optimal Solution

Approach

We need to turn a tree into a simple text string and then rebuild the tree from that string. The key is to serialize the tree in a way that preserves its structure, and then use that structure to reconstruct it exactly.

Here's how the algorithm would work step-by-step:

First, we walk through the tree in a specific order, like going down each branch level by level.
As we visit each node, we write its value into a string. If a node is missing (empty), we write a special symbol like 'X' to mark that spot.
We also add separators, like commas, between the node values so we can tell where each value begins and ends when we read the string back.
Now, to rebuild the tree, we take our string and split it back into the individual values using the separators.
We then use these values to recreate the tree level by level, using the 'X' symbols to tell us where branches are missing.
Because we visited the tree in a consistent order during serialization, we can reconstruct it perfectly as long as we do the inverse process correctly.

Code Implementation

class TreeNode:
    def __init__(self, value):
        self.value = value
        self.left = None
        self.right = None

class Codec:
    def serialize(self, root):
        if not root:
            return ''

        serialized_tree_string = ''
        queue = [root]

        while queue:
            current_node = queue.pop(0)
            
            if current_node:
                serialized_tree_string += str(current_node.value) + ','
                queue.append(current_node.left)
                queue.append(current_node.right)
            else:
                serialized_tree_string += 'X,'

        return serialized_tree_string

    def deserialize(self, data):
        if not data:
            return None

        values = data.split(',')
        values.pop()
        
        root_value = values.pop(0)
        if root_value == 'X':
            return None

        root = TreeNode(int(root_value))
        queue = [root]

        # Reconstruct the tree level by level
        while values:
            node = queue.pop(0)
            
            left_value = values.pop(0)

            # 'X' represents a null node.
            if left_value != 'X':
                node.left = TreeNode(int(left_value))
                queue.append(node.left)

            right_value = values.pop(0)

            # 'X' represents a null node.
            if right_value != 'X':
                node.right = TreeNode(int(right_value))
                queue.append(node.right)

        return root

Big(O) Analysis

Time Complexity

O(n) – Serializing the BST involves traversing all n nodes in the tree exactly once, adding each node's value (or 'X' for null nodes) to a string. Deserializing also processes each of the n values in the serialized string once to reconstruct the tree. Therefore, both serialization and deserialization have a time complexity directly proportional to the number of nodes n, resulting in O(n).

Space Complexity

O(N) – The serialization process implicitly uses a queue (or list) to perform a level-order traversal. In the worst-case scenario, like a complete binary tree, the queue will hold roughly half of the nodes at the deepest level before they are processed. This translates to storing approximately N/2 nodes in the queue where N is the total number of nodes in the tree. Therefore, the auxiliary space required by the queue is proportional to N, resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
Null or empty BST root	Serialize to empty string and deserialize to null.
BST with only one node	Serialize to a string representing the single node's value; deserialize creates a BST with that single node.
Completely unbalanced BST (left-skewed or right-skewed)	Serialization will still process each node and Deserialization will reconstruct the skewed tree, albeit potentially inefficiently due to recursion depth.
BST with duplicate values	The serialization method should consistently represent the tree structure, and deserialization rebuilds the structure including duplicates where present.
BST containing negative numbers, zero, and positive numbers	Serialization and deserialization should handle all integer values within the representable range without modification.
Very large BST (potentially exceeding memory limits)	Consider using an iterative approach to serialization/deserialization, or a different serialization format to reduce memory overhead, while being mindful of potential stack overflow errors if using recursion.
Integer overflow during serialization/deserialization when converting to/from strings	Use appropriate data types and error handling to prevent integer overflow or underflow during string conversions.
Different serialization methods producing different, but valid, BSTs upon deserialization.	The solution should ensure that any serialization method, when deserialized, results in a BST with the same values and structural properties, even if not identical in node object identity.

Null or empty BST root

How to Handle:

Serialize to empty string and deserialize to null.

BST with only one node

How to Handle:

Serialize to a string representing the single node's value; deserialize creates a BST with that single node.

Completely unbalanced BST (left-skewed or right-skewed)

How to Handle:

Serialization will still process each node and Deserialization will reconstruct the skewed tree, albeit potentially inefficiently due to recursion depth.

BST with duplicate values

How to Handle:

The serialization method should consistently represent the tree structure, and deserialization rebuilds the structure including duplicates where present.

BST containing negative numbers, zero, and positive numbers

How to Handle:

Serialization and deserialization should handle all integer values within the representable range without modification.

Very large BST (potentially exceeding memory limits)

How to Handle:

Consider using an iterative approach to serialization/deserialization, or a different serialization format to reduce memory overhead, while being mindful of potential stack overflow errors if using recursion.

Integer overflow during serialization/deserialization when converting to/from strings

How to Handle:

Use appropriate data types and error handling to prevent integer overflow or underflow during string conversions.

Different serialization methods producing different, but valid, BSTs upon deserialization.

How to Handle:

The solution should ensure that any serialization method, when deserialized, results in a BST with the same values and structural properties, even if not identical in node object identity.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What data type are the node values? Are negative values allowed?
Can the BST be empty (null root)?
Does the serialization need to be space-optimal, or is readability/simplicity prioritized?
If there are multiple valid BSTs that can be constructed from the serialized string, is any valid BST acceptable upon deserialization?
Are there any constraints on the depth or number of nodes in the BST?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start by exploring representing the tree node by node.
Consider different ways to represent the value of each node and whether the node has children.
Try all the possible orders to list out each of the node values.
For each order, check if it is a valid ordering to reconstruct the tree.
Consider different ways to represent null or non-existent nodes.
Keep track of all the valid representations.
Among all the valid representations, pick one as the final serialized string.
To deserialize, we'd then attempt to rebuild the BST from each possible structure represented by the string, making sure it is actually valid.

Code Implementation

class TreeNode:
    def __init__(self, value):
        self.value = value
        self.left = None
        self.right = None

class Solution:
    def serialize(self, root):
        # Brute force serialization - generate all possibilities
        if not root:
            return ""

        nodes = []
        def traverse(node):
            if not node:
                nodes.append(None)
                return
            nodes.append(node.value)
            traverse(node.left)
            traverse(node.right)

        traverse(root)

        # Representation of all possible node orderings as string
        return ','.join(map(str, [node if node is None else node for node in nodes]))

    def deserialize(self, data):
        # Brute force deserialization - try all tree structures
        if not data:
            return None

        values = data.split(',')
        self.index = 0

        def build_tree():
            # Reconstruct the tree using recursive calls
            if self.index >= len(values) or values[self.index] == 'None':
                self.index += 1
                return None

            current_value = int(values[self.index])
            self.index += 1
            node = TreeNode(current_value)

            # Recursively build left and right subtrees
            node.left = build_tree()

            # Guaranteeing correct BST properties is crucial
            node.right = build_tree()
            return node

        return build_tree()

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

First, we walk through the tree in a specific order, like going down each branch level by level.
As we visit each node, we write its value into a string. If a node is missing (empty), we write a special symbol like 'X' to mark that spot.
We also add separators, like commas, between the node values so we can tell where each value begins and ends when we read the string back.
Now, to rebuild the tree, we take our string and split it back into the individual values using the separators.
We then use these values to recreate the tree level by level, using the 'X' symbols to tell us where branches are missing.
Because we visited the tree in a consistent order during serialization, we can reconstruct it perfectly as long as we do the inverse process correctly.

Code Implementation

class TreeNode:
    def __init__(self, value):
        self.value = value
        self.left = None
        self.right = None

class Codec:
    def serialize(self, root):
        if not root:
            return ''

        serialized_tree_string = ''
        queue = [root]

        while queue:
            current_node = queue.pop(0)
            
            if current_node:
                serialized_tree_string += str(current_node.value) + ','
                queue.append(current_node.left)
                queue.append(current_node.right)
            else:
                serialized_tree_string += 'X,'

        return serialized_tree_string

    def deserialize(self, data):
        if not data:
            return None

        values = data.split(',')
        values.pop()
        
        root_value = values.pop(0)
        if root_value == 'X':
            return None

        root = TreeNode(int(root_value))
        queue = [root]

        # Reconstruct the tree level by level
        while values:
            node = queue.pop(0)
            
            left_value = values.pop(0)

            # 'X' represents a null node.
            if left_value != 'X':
                node.left = TreeNode(int(left_value))
                queue.append(node.left)

            right_value = values.pop(0)

            # 'X' represents a null node.
            if right_value != 'X':
                node.right = TreeNode(int(right_value))
                queue.append(node.right)

        return root

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty BST root	Serialize to empty string and deserialize to null.
BST with only one node	Serialize to a string representing the single node's value; deserialize creates a BST with that single node.
Completely unbalanced BST (left-skewed or right-skewed)	Serialization will still process each node and Deserialization will reconstruct the skewed tree, albeit potentially inefficiently due to recursion depth.
BST with duplicate values	The serialization method should consistently represent the tree structure, and deserialization rebuilds the structure including duplicates where present.
BST containing negative numbers, zero, and positive numbers	Serialization and deserialization should handle all integer values within the representable range without modification.
Very large BST (potentially exceeding memory limits)	Consider using an iterative approach to serialization/deserialization, or a different serialization format to reduce memory overhead, while being mindful of potential stack overflow errors if using recursion.
Integer overflow during serialization/deserialization when converting to/from strings	Use appropriate data types and error handling to prevent integer overflow or underflow during string conversions.
Different serialization methods producing different, but valid, BSTs upon deserialization.	The solution should ensure that any serialization method, when deserialized, results in a BST with the same values and structural properties, even if not identical in node object identity.

Null or empty BST root

How to Handle:

Serialize to empty string and deserialize to null.

BST with only one node

How to Handle:

Serialize to a string representing the single node's value; deserialize creates a BST with that single node.

Completely unbalanced BST (left-skewed or right-skewed)

How to Handle:

Serialization will still process each node and Deserialization will reconstruct the skewed tree, albeit potentially inefficiently due to recursion depth.

BST with duplicate values

How to Handle:

The serialization method should consistently represent the tree structure, and deserialization rebuilds the structure including duplicates where present.

BST containing negative numbers, zero, and positive numbers

How to Handle:

Serialization and deserialization should handle all integer values within the representable range without modification.

Very large BST (potentially exceeding memory limits)

How to Handle:

Integer overflow during serialization/deserialization when converting to/from strings

How to Handle:

Use appropriate data types and error handling to prevent integer overflow or underflow during string conversions.

Different serialization methods producing different, but valid, BSTs upon deserialization.

How to Handle:

The solution should ensure that any serialization method, when deserialized, results in a BST with the same values and structural properties, even if not identical in node object identity.

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