# Unique Binary Search Trees Leetcode Solution

In this post, we are going to solve the Unique Binary Search Trees Leetcode SolutionÂ problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

## Problem

Given an integerÂ `n`, returnÂ the number of structurally uniqueÂ BSTâ€™s (binary search trees) which has exactlyÂ `n`Â nodes of unique values fromÂ `1`Â toÂ `n`.

Example 1:

```Input: n = 3
Output: 5
```

Example 2:

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

Constraints:

• `1 <= n <= 19`

Now, letâ€™s see the leetcode solution ofÂ Unique Binary Search Trees Leetcode Solution.

### Unique Binary Search Trees Leetcode 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]
```

### Unique Binary Search Trees Leetcode Solutionin 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[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 Leetcode 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];
}
}
```

Note:Â This problemÂ Unique Binary Search Trees is generated byÂ LeetcodeÂ but the solution is provided byÂ Chase2learn This tutorial is only forÂ EducationalÂ andÂ LearningÂ purposes.

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