What is the time complexity of Unique Binary Search Trees II?

The optimal solution has a time complexity of O(4^n/sqrt(n)).

What is the space complexity of Unique Binary Search Trees II?

The space complexity is O(4^n/sqrt(n)).

What data structures are used to solve Unique Binary Search Trees II?

This problem uses Dynamic Programming, Backtracking, Tree, Binary Search Tree, Binary Tree. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Unique Binary Search Trees II?

Insight 1: The number of unique BSTs grows rapidly with `n`. Brute force generation of every possible tree would be inefficient. Insight 2: Dynamic programming with memoization is crucial. The problem can be broken down into smaller subproblems: generating all possible left subtrees and right subtrees for a given root.

Which companies ask LeetCode 95. Unique Binary Search Trees II in interviews?

This problem is frequently asked at Amazon, Apple, Bloomberg, Meta, Microsoft and 1 more companies.

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Unique Binary Search Trees II - LeetCode 95 Solution

Q: How to solve Unique Binary Search Trees II?

The provided solution uses a recursive approach with memoization (dynamic programming) to efficiently generate all unique BSTs. The core idea is to iterate through all possible root values (from `start` to `end`) and then recursively generate all possible left subtrees (from `start` to `i-1`) and right subtrees (from `i+1` to `end`). Then, combine each possible left subtree with each possible right subtree to form a new BST with root `i`. Memoization is used to store the results of already computed subproblems to avoid redundant calculations. This dramatically improves performance.

Unique Binary Search Trees II - Complete Solution Guide

Unique Binary Search Trees II is LeetCode problem 95, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an integer n , return all the structurally unique BST' s (binary search trees), which has exactly n nodes of unique values from 1 to n . Return the answer in any order . Example 1: Input: n = 3 Output: [[1,null,2,null,3],[1,null,3,2],[2,1,3],[3,1,null,null,2],[3,2,null,1]] Example 2: Input: n = 1 Output: [[1]] Constraints: 1 <= n <= 8

Detailed Explanation

The problem asks us to generate all structurally unique Binary Search Trees (BSTs) for a given number `n`. These BSTs must contain nodes with unique values from 1 to `n`. The output should be a list containing the root nodes of all possible BSTs.

Solution Approach

The provided solution uses a recursive approach with memoization (dynamic programming) to efficiently generate all unique BSTs. The core idea is to iterate through all possible root values (from `start` to `end`) and then recursively generate all possible left subtrees (from `start` to `i-1`) and right subtrees (from `i+1` to `end`). Then, combine each possible left subtree with each possible right subtree to form a new BST with root `i`. Memoization is used to store the results of already computed subproblems to avoid redundant calculations. This dramatically improves performance.

Step-by-Step Algorithm

Step 1: Define a recursive function `build(start, end)` that takes a start and end value, representing the range of numbers to build BSTs from.
Step 2: Base Case: If `start > end`, it means there are no numbers in the range, so return a list containing only `None` (representing an empty subtree). This signifies the end of a branch.
Step 3: Memoization: Before computing, check if the result for `(start, end)` is already in the memo. If so, return the memoized result.
Step 4: Iteration: Iterate through all possible root values `i` from `start` to `end`.
Step 5: Recursive Calls: For each `i`, recursively call `build(start, i-1)` to get all possible left subtrees and `build(i+1, end)` to get all possible right subtrees.
Step 6: Tree Construction: Iterate through all possible combinations of left and right subtrees. Create a new `TreeNode` with value `i`, set the left child to the current left subtree and the right child to the current right subtree.
Step 7: Accumulate Results: Add the newly created tree to the result list.
Step 8: Memoize Result: After processing all possible root values, store the result list in the memo for future use.
Step 9: Return Result: Return the result list.

Key Insights

Insight 1: The number of unique BSTs grows rapidly with `n`. Brute force generation of every possible tree would be inefficient.
Insight 2: Dynamic programming with memoization is crucial. The problem can be broken down into smaller subproblems: generating all possible left subtrees and right subtrees for a given root.
Insight 3: Recursive approach is natural for generating trees. We can iterate through each number from 1 to n, treat it as the root, recursively construct all possible left and right subtrees and combine them

Complexity Analysis

Time Complexity: O(4^n/sqrt(n))

Space Complexity: O(4^n/sqrt(n))

Topics

This problem involves: Dynamic Programming, Backtracking, Tree, Binary Search Tree, Binary Tree.

Companies

Asked at: Amazon, Apple, Bloomberg, Meta, Microsoft, Uber.

Unique Binary Search Trees II | LeetCode 95

Medium

Dynamic Programming Backtracking Tree Binary Search Tree Binary Tree

Time: O(4^n/sqrt(n))Space: O(4^n/sqrt(n))

Solution Code

Available in:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<TreeNode> generateTrees(int n) {
        return build(1, n);
    }

    private HashMap<String, List<TreeNode>> memo = new HashMap<>();

    private List<TreeNode> build(int start, int end) {
        if (start > end) {
            List<TreeNode> list = new ArrayList<>();
            list.add(null);
            return list;
        }

        String key = start + "," + end;
        if (memo.containsKey(key)) {
            return memo.get(key);
        }

        List<TreeNode> result = new ArrayList<>();
        for (int i = start; i <= end; i++) {
            List<TreeNode> leftSubtrees = build(start, i - 1);
            List<TreeNode> rightSubtrees = build(i + 1, end);

            for (TreeNode leftNode : leftSubtrees) {
                for (TreeNode rightNode : rightSubtrees) {
                    TreeNode root = new TreeNode(i, leftNode, rightNode);
                    result.add(root);
                }
            }
        }

        memo.put(key, result);
        return result;
    }
}

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Problem Info

Description

Problem 95: Unique Binary Search Trees II - Detailed explanation and solution approaches.

Topics

Dynamic Programming Backtracking Tree Binary Search Tree Binary Tree

Companies

AmazonAppleBloombergMetaMicrosoftUber

Stats

Difficulty:Medium

Languages:4

Companies:6

Problem #:95

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Define a recursive function `build(start, end)` that takes a start and end value, representing the range of numbers to build BSTs from.
2
Step 2: Base Case: If `start > end`, it means there are no numbers in the range, so return a list containing only `None` (representing an empty subtree). This signifies the end of a branch.
3
Step 3: Memoization: Before computing, check if the result for `(start, end)` is already in the memo. If so, return the memoized result.
4
Step 4: Iteration: Iterate through all possible root values `i` from `start` to `end`.
5
Step 5: Recursive Calls: For each `i`, recursively call `build(start, i-1)` to get all possible left subtrees and `build(i+1, end)` to get all possible right subtrees.
6
Step 6: Tree Construction: Iterate through all possible combinations of left and right subtrees. Create a new `TreeNode` with value `i`, set the left child to the current left subtree and the right child to the current right subtree.
7
Step 7: Accumulate Results: Add the newly created tree to the result list.
8
Step 8: Memoize Result: After processing all possible root values, store the result list in the memo for future use.
9
Step 9: Return Result: Return the result list.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(4^n/sqrt(n))

💾

Space Complexity

O(4^n/sqrt(n))

📝

Why?

The time complexity is dictated by the recursive calls and the Catalan number. The memoization avoids redundant computation, but the initial computation still requires generating a number of BSTs comparable to the Catalan number for n. The space complexity stems from storing the results of these subproblems in the memoization table. Both time and space complexities are directly related to the Catalan number's growth.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Iterative Dynamic Programming

An iterative DP approach can also be used to solve this problem, though it's typically more complex to implement and understand than the recursive approach. You'd need to build up a table of all possible BSTs for ranges of increasing size.

Time: O(4^n/sqrt(n))Space: O(4^n/sqrt(n))

Trade-offs: The iterative approach avoids recursion overhead, which can be beneficial for larger `n`. However, it often leads to more complex and harder-to-read code. It might save a small amount of stack space, but will still have the Catalan number dominated space complexity. The recursive memoized approach is generally preferred for its clarity.

✨

Key Insights

1Insight 1: The number of unique BSTs grows rapidly with `n`. Brute force generation of every possible tree would be inefficient.
2Insight 2: Dynamic programming with memoization is crucial. The problem can be broken down into smaller subproblems: generating all possible left subtrees and right subtrees for a given root.
3Insight 3: Recursive approach is natural for generating trees. We can iterate through each number from 1 to n, treat it as the root, recursively construct all possible left and right subtrees and combine them

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: n = 0 or start > end. Handled correctly by returning a list containing only null (representing an empty tree). This is critical for the recursion to terminate properly when there are no nodes to generate a tree from.
Edge case 2: n = 1. The algorithm correctly generates a single BST with one node.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using memoization. Without memoization, the algorithm will recalculate the same subproblems many times, resulting in exponential time complexity and causing timeouts for larger values of `n`.

✗Common mistake 2: Incorrect base case. Failing to handle the case where `start > end` correctly will lead to incorrect results or infinite recursion.

✗Common mistake 3: Forgetting to free the allocated memory in C. Memory leaks can cause problems over time if not properly managed.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Compiler design. BSTs are used to represent the syntax of programming languages. Generating all possible parse trees for a given code snippet could be a simplified example of this problem.

🏦Real-world scenario 2: Database indexing. BSTs and their variations (like balanced BSTs) are fundamental data structures for indexing in databases. Understanding the different possible tree structures is useful for optimizing database performance.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 4, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

How do I avoid duplicate solutions in backtracking?

Sort the input first, then skip consecutive duplicates at the same recursion level: if (i > start && nums[i] == nums[i-1]) continue. For permutations, use a visited array or swap-based approach. Always consider if order matters (permutations vs combinations).

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Unique Binary Search Trees II"?

For this medium-level Dynamic Programming problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether backtracking techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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