Unique Binary Search Trees II
Given an integer $n$, return all structurally unique Binary Search Trees (BSTs) which store values 1 to $n$. Use Recursion/DP.
Why Interviewers Ask This
Meta evaluates this question to assess a candidate's mastery of recursive backtracking and dynamic programming optimization. They specifically look for the ability to handle state explosion in combinatorial problems while maintaining clean, readable code. The problem tests if you can decompose a complex structural generation task into smaller subproblems involving root selection and left/right subtree construction.
How to Answer This Question
1. Clarify the problem constraints and confirm that values 1 to n must form a valid BST where inorder traversal yields sorted order. 2. Propose a recursive strategy: iterate through each number i from start to end, treating it as the root. 3. Explain how to generate all possible left subtrees using numbers less than i and all right subtrees using numbers greater than i. 4. Combine every unique left subtree with every unique right subtree to form complete trees rooted at i. 5. Discuss memoization (DP) to cache results for ranges [start, end], preventing redundant calculations for identical sub-ranges. 6. Conclude by analyzing time complexity O(4^n / sqrt(n)) and space complexity, showing awareness of Meta's focus on efficient algorithmic design.
Key Points to Cover
- Explicitly stating the relationship between the chosen root and the partitioning of remaining values into left and right sets
- Demonstrating the Cartesian product logic for combining independent left and right subtree lists
- Implementing memoization to avoid recalculating identical sub-range results
- Correctly handling base cases where the start index exceeds the end index
- Articulating the exponential time complexity related to Catalan numbers
Sample Answer
To solve Unique Binary Search Trees II, I first recognize that for any range of numbers, choosing a specific number as the root dictates the structure of its left and right children based on BST properties. My approach uses recursion with memoization. For a given range from start to end, I iterate through each number i acting as the potential root. Then, I recursively generate all unique left subtrees using the range [start, i-1] and all unique right subtrees using [i+1, end]. Once these lists are generated, I perform a Cartesian product to pair every left subtree with every right subtree, attaching them to the new root node i. To optimize, I will use a map to store results for each range pair, ensuring we don't recompute the same sub-problems multiple times. For example, if n is 3, choosing 2 as root requires combining left subtrees from {1} with right subtrees from {3}. This method ensures we cover all structural possibilities without duplicates. I would implement this in Python or C++ using a helper function that returns a list of tree nodes. Finally, I'd discuss the Catalan number growth rate to demonstrate understanding of why this problem becomes computationally expensive quickly, which aligns with Meta's emphasis on scalable solutions.
Common Mistakes to Avoid
- Failing to handle the empty range case correctly, leading to null pointer exceptions when building leaf nodes
- Not using memoization, resulting in Time Limit Exceeded errors due to exponential recomputation
- Confusing value uniqueness with structural uniqueness, potentially generating duplicate tree structures
- Ignoring the BST property that left children must be smaller and right children larger than the root
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