Verify Preorder Serialization of a Binary Tree

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TreesStacksStrings

One way to serialize a binary tree is to use preorder traversal. When we encounter a non-null node, we record the node's value. If it is a null node, we record using a sentinel value such as '#'.

For example, the above binary tree can be serialized to the string "9,3,4,#,#,1,#,#,2,#,6,#,#", where '#' represents a null node.

Given a string of comma-separated values preorder, return true if it is a correct preorder traversal serialization of a binary tree.

It is guaranteed that each comma-separated value in the string must be either an integer or a character '#' representing null pointer.

You may assume that the input format is always valid.

For example, it could never contain two consecutive commas, such as "1,,3".

Note: You are not allowed to reconstruct the tree.

Example 1:

Input: preorder = "9,3,4,#,#,1,#,#,2,#,6,#,#"
Output: true

Example 2:

Input: preorder = "1,#"
Output: false

Example 3:

Input: preorder = "9,#,#,1"
Output: false

Constraints:

1 <= preorder.length <= 10⁴
preorder consist of integers in the range [0, 100] and '#' separated by commas ','.

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:

Can the nodes in the binary tree have any integer value, or are they restricted to single digits or a specific range?
Can the input string be empty or null? What should I return in those cases?
Does 'X' strictly represent a null node, or could there be other characters representing null?
Is the input string guaranteed to be a valid preorder serialization of *some* tree structure (possibly incomplete), or could it be completely invalid?
Could the preorder traversal contain only 'X's, representing an empty tree?

Brute Force Solution

Approach

Think of building a tree step-by-step. The brute force method explores every single possible tree structure that could be made from the given sequence, and checks if any of them match the rules of a valid binary tree serialization. This involves trying all combinations and arrangements.

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

Start at the very beginning of the sequence.
Try to interpret the first element as the root of the tree.
Based on what the root node looks like (whether it's a value or a null marker), decide if you need to look for children.
If children are needed, try every possible way to assign the next elements in the sequence as the left and right children.
For each child, treat that child as a new root and recursively apply the same process: trying every combination of its descendants.
Keep going until you've used up the entire sequence, or you find a combination that doesn't work (violates the tree structure rules).
If you get to the end of the sequence and the tree is valid, you've found a correct serialization. If not, go back and try a different combination earlier in the process.
Continue exploring all possible combinations until you either find a valid serialization or exhaust all possibilities, determining that the sequence cannot represent a valid binary tree serialization.

Code Implementation

def verify_preorder_brute_force(preorder):    preorder_list = preorder.split(',')
    def is_valid_serialization_recursive(preorder_list, index):        # Base case: if we've reached the end, check for completion
        if index == len(preorder_list):            return False, index
        
        root_value = preorder_list[index]
        
        # Check if current node is null
        if root_value == '#':            return True, index + 1
        
        # Attempt to construct the left subtree
        is_left_valid, next_index = is_valid_serialization_recursive(preorder_list, index + 1)
        
        if not is_left_valid:
            return False, next_index
        
        # Attempt to construct the right subtree
        is_right_valid, final_index = is_valid_serialization_recursive(preorder_list, next_index)
        
        if not is_right_valid:
            return False, final_index
        
        return True, final_index
    
    is_valid, final_index = is_valid_serialization_recursive(preorder_list, 0)
    # Check if we've consumed the entire preorder string and the tree is valid
    return is_valid and final_index == len(preorder_list)

Big(O) Analysis

Time Complexity

O(exponential) – The brute force approach explores every possible binary tree structure that can be formed from the input string. The number of possible binary trees with n nodes grows exponentially. For each node, the algorithm explores possibilities of assigning children and recursively calls itself. Thus the time complexity reflects the combinatorial nature of exploring all possible tree structures leading to an exponential time complexity, where the base of the exponent is significantly greater than 2 and influenced by the frequency of non-null nodes, making a precise formula intractable.

Space Complexity

O(N) – The described brute force approach explores all possible tree structures using recursion. Each recursive call creates a new stack frame to manage the state of exploring a subtree. In the worst-case scenario, where the input represents a highly skewed tree, the maximum depth of the recursion can reach N, where N is the number of nodes in the input sequence. Thus, the auxiliary space required for the recursion stack grows linearly with the input size N. Therefore, the overall space complexity is O(N).

Optimal Solution

Approach

The key idea is to think about the 'capacity' or 'slots' available as we traverse the preorder string. Each number needs a slot, and each null node frees up a slot. We efficiently track this balance to determine validity without building the tree.

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

Imagine you're building the tree and have a certain number of 'empty slots' available.
Start with one initial slot available since a root node is needed to start building.
As you go through each element in the string, do the following:
If you find a number (not a null), it uses up one available slot but opens two more slots (for its potential left and right children). So, decrease available slots by one and increase by two; effectively, increase by one.
If you find a null, it uses up one available slot but doesn't open any new slots. So, just decrease available slots by one.
If, at any point, your available slots drop to zero *before* you've reached the end of the string, it means you've built a complete subtree and are trying to add nodes to nowhere. Thus the string is invalid.
After processing the entire string, the serialization is valid if and only if the number of available slots is exactly zero.

Code Implementation

def is_valid_serialization(preorder): 
    available_slots = 1
    nodes = preorder.split(',')

    for node in nodes:
        # If no slots are available, the string is invalid.
        if available_slots == 0:
            return False

        available_slots -= 1

        # Null nodes do not add any more slots.
        if node != '#':
            # Add two slots for potential children.
            available_slots += 2

    # The serialization is valid if all slots are used.
    return available_slots == 0

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the preorder string exactly once. The input size, n, represents the number of nodes in the serialized tree (including null nodes). Each node is processed in constant time by updating the 'slots' counter. Therefore, the time complexity is directly proportional to the number of nodes in the string, resulting in O(n) time complexity.

Space Complexity

O(1) – The provided algorithm uses a constant amount of extra space. It only maintains a single integer variable representing the number of available slots. This variable's memory footprint remains constant regardless of the size of the input string, denoted as N. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty string input	Return true if the string is empty, as an empty tree can be serialized as ''.
Single '#' representing a null node	Return true, as a single null node is a valid, complete serialization of an empty tree.
String with just one number	Return false because a single node must have children (even if they are null).
String starting with '#'	Return false because the root cannot be null.
String ending with a number	Return false because there must be enough null nodes at the end to complete the tree.
Deeply unbalanced tree serialization	The stack-based or slot-based approach should handle this efficiently without excessive recursion depth.
Serialization of a complete binary tree	The algorithm correctly identifies complete trees and returns true.
Serialization with numbers containing multiple digits	The string splitting logic must correctly parse multi-digit numbers, like '9,3,4,#,#,1,#,#,2,#,6,#,#'.

Null or empty string input

How to Handle:

Return true if the string is empty, as an empty tree can be serialized as ''.

Single '#' representing a null node

How to Handle:

Return true, as a single null node is a valid, complete serialization of an empty tree.

String with just one number

How to Handle:

Return false because a single node must have children (even if they are null).

String starting with '#'

How to Handle:

Return false because the root cannot be null.

String ending with a number

How to Handle:

Return false because there must be enough null nodes at the end to complete the tree.

Deeply unbalanced tree serialization

How to Handle:

The stack-based or slot-based approach should handle this efficiently without excessive recursion depth.

Serialization of a complete binary tree

How to Handle:

The algorithm correctly identifies complete trees and returns true.

Serialization with numbers containing multiple digits

How to Handle:

The string splitting logic must correctly parse multi-digit numbers, like '9,3,4,#,#,1,#,#,2,#,6,#,#'.