How do you solve Verify Preorder Serialization of a Binary Tree on LeetCode?

Verify Preorder Serialization of a Binary Tree (LeetCode #331) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Each non-null node creates two new slots, while each null node consumes one slot.

What is the time complexity of Verify Preorder Serialization of a Binary Tree?

The optimal solution for Verify Preorder Serialization of a Binary Tree runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only pass through the list of nodes once, and the space complexity is O(n) due to the storage of the split nodes.

What is the brute force solution for Verify Preorder Serialization of a Binary Tree?

The brute force approach for Verify Preorder Serialization of a Binary Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. We can simulate the process of constructing the binary tree from the preorder serialization. For each node, we will check if we can create a valid tree structure by counting the number of nodes and nulls we encounter. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Verify Preorder Serialization of a Binary Tree?

Verify Preorder Serialization of a Binary Tree uses the following concepts: String, Stack, Tree, Binary Tree. The recommended patterns to study are: Greedy, Tree Traversal.

What are the interview tips for Verify Preorder Serialization of a Binary Tree?

Explain your thought process clearly while coding. Make sure to handle edge cases, such as a single null node. Practice dry-running your code with sample inputs before finalizing.

What are common mistakes when solving Verify Preorder Serialization of a Binary Tree?

Forgetting to check if slots go negative during iteration. Not handling the case where the final slot count is not zero.

#331

Verify Preorder Serialization of a Binary Tree

Medium

StringStackTreeBinary TreeGreedyTree Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

We can achieve a more efficient solution by using a single pass through the nodes while maintaining a count of available slots for new nodes, which allows us to validate the preorder serialization in linear time.

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Algorithm

6 steps

1Step 1: Split the input string by commas to get a list of nodes.
2Step 2: Initialize a counter for available slots, starting with 1.
3Step 3: For each node in the list, decrement the slot count for the current node.
4Step 4: If the node is not a '#', increment the slot count by 2 (since it creates two new slots).
5Step 5: If at any point the slot count goes negative, return false.
6Step 6: After processing all nodes, return true if the slot count is zero.

solution.py10 lines

1def isValidSerialization(preorder):
2    nodes = preorder.split(',')
3    slots = 1
4    for node in nodes:
5        slots -= 1
6        if slots < 0:
7            return False
8        if node != '#':
9            slots += 2
10    return slots == 0

ℹ

Complexity note: The time complexity is O(n) because we only pass through the list of nodes once, and the space complexity is O(n) due to the storage of the split nodes.

1Each non-null node creates two new slots, while each null node consumes one slot.
2The final count of slots must be zero for a valid serialization.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.