# Unique Binary Search Trees

Published: Jan 17, 2023

Medium Dynamic Programming Math Binary Search Tree

## Problem Description

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`.

Constraints:

• `1 <= n <= 19`

https://leetcode.com/problems/unique-binary-search-trees/

## Examples

``````Example 1
Input: n = 3
Output: 5
``````
``````Exmaple 2
Input: n = 1
Output: 1
``````

## How to Solve

This problem asks the Catalan number. The Catalan number is from combinatorial mathematics. The sequence of natural numbers are: 1, 1, 2, 5, 14, 42, 132, 249, 1430, 4862, … To calculate the Catalan number, we can use a mathematical formula.

``````C[0] = 1
C[n + 1] = (2 * (2n + 1) / n + 2) * C[n]
``````

Looping over the formula from 0 to n will give us the answer.

Aside of the mathematical formula, the problem can be solved by dynamic programming (DP). The formula above says the next state is depends on the previous state. It means the DP is a good approach. The auxiliary array saves the Catalan numbers from 0 to n, which is calculated as:

``````C[0] = 1
C[n + 1] = sum(C[i] * C[n - i]) where i = 0 to n
``````

Above relation might be easier to understand compared to the mathematical formula.

The solutions here show both mathematical and DP approaches.

For the Catalan number, see Catalan number for details.

## Solution

``````class UniqueBinarySearchTrees {
public:
int numTreesMath(int n) {
long C = 1;
for (int i = 0; i < n; ++i)
{
C = C * 2 * (2 * i + 1) / (i + 2);
}
return C;
}

int numTreesDP(int n) {
vector<long long int> memo(n + 1, 0);
memo[0] = 1, memo[1] = 1;
for (int i = 2; i <= n; ++i) {
for (int j = 0; j < i; ++j) {
memo[i] += memo[j] * memo[i - j - 1];
}
}
return (int)memo[n];
}
};
``````
``````class UniqueBinarySearchTrees {
public int numTreesMaath(int n) {
long C = 1;
for (int i = 0; i < n; ++i)
{
C = C * 2 * (2 * i + 1) / (i + 2);
}
return (int)C;
}

public int numTreesDP(int n) {
long[] memo = new long[n+1];
memo[0] = memo[1] = 1;
for (int i = 2; i <= n; i++) {
for (int j = 0; j < i; j++) {
memo[i] += memo[j] * memo[i-j-1];
}
}
return (int)memo[n];
}
}

``````
``````
``````
``````class UniqueBinarySearchTrees:
def numTreesMath(self, n: int) -> int:
C = 1
for i in range(n):
C = C * 2 * (2 * i + 1) // (i + 2)
return C

def numTreesDP(self, n: int) -> int:
memo = [0 for _ in range(n + 1)]
memo[0], memo[1] = 1, 1
for i in range(2, n + 1):
for j in range(1, i + 1):
memo[i] += memo[j - 1] * memo[i - j]
return memo[n]
``````
``````
``````

## Complexities

• Math
• Time: `O(n)`
• Space: `O(1)`
• DP
• Time: `O(n^2)`
• Space: `O(n)`