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In this post, you will find the solution for the **Unique Binary Search Trees** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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LeetCode has over 1,900 questions for you to practice, covering many different programming concepts. Every coding problem has a classification of either *Easy*, *Medium*, or *Hard*.

**LeetCode problems focus on algorithms and data structures. Here is some topic you can find problems on LeetCode:**

- Mathematics/Basic Logical Based Questions
- Arrays
- Strings
- Hash Table
- Dynamic Programming
- Stack & Queue
- Trees & Graphs
- Greedy Algorithms
- Breadth-First Search
- Depth-First Search
- Sorting & Searching
- BST (Binary Search Tree)
- Database
- Linked List
- Recursion, etc.

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** Link for the Problem** – Unique Binary Search Trees– LeetCode Problem

Unique Binary Search Trees– LeetCode Problem

**Problem:**

Given an integer `n`

, return *the number of structurally unique BST’s (binary search trees) which has exactly *

`n`

*nodes of unique values from*

`1`

*to*

`n`

.**Example 1:**

Input:n = 3Output:5

**Example 2:**

Input:n = 1Output:1

**Constraints:**

Unique Binary Search Trees– LeetCode Solutions

Unique Binary Search Trees Solution in C++:

class Solution { public: int numTrees(int n) { // G[i] := # of unique BST's that store values 1..i vectorG(n + 1); G[0] = 1; G[1] = 1; for (int i = 2; i <= n; ++i) for (int j = 0; j < i; ++j) G[i] += G[j] * G[i - j - 1]; return G[n]; } };

Unique Binary Search Trees Solution in Java:

class Solution { public int numTrees(int n) { // G[i] := # of unique BST's that store values 1..i int[] G = new int[n + 1]; G[0] = 1; G[1] = 1; for (int i = 2; i <= n; ++i) for (int j = 0; j < i; ++j) G[i] += G[j] * G[i - j - 1]; return G[n]; } }

Unique Binary Search Trees Solution in Python:

class Solution: def numTrees(self, n: int) -> int: # G[i] := # of unique BST's that store values 1..i G = [1, 1] + [0] * (n - 1) for i in range(2, n + 1): for j in range(i): G[i] += G[j] * G[i - j - 1] return G[n]