# Unique Binary Search Trees Leetcode Solution In this post, you will know how to solve the Unique Binary Search Trees Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Leetcode by yourself. If you find any difficulty after trying several times, then you can look for solutions.

## 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 = 3
Output: 5
```

Example 2:

```Input: n = 1
Output: 1
```

Constraints:

• `1 <= n <= 19`

### Unique Binary Search Trees Leetcode Solutions in Python

```class Solution:
def numTrees(self, n: int) -> int:
# G[i] := # Of unique BST's that store values 1..i
G = [1, 1] +  * (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]
```

### Unique Binary Search Trees Leetcode Solutions in CPP

```class Solution {
public:
int numTrees(int n) {
// G[i] := # of unique BST's that store values 1..i
vector<int> G(n + 1);
G = 1;
G = 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 Leetcode Solutions 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 = 1;
G = 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];
}
}
```

Note: This problem Unique Binary Search Trees is generated by Leetcode but the solution is provided by Brokenprogrammers. This tutorial is only for Educational and Learning purposes.