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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 <= 19We define $f[i]$ to represent the number of binary search trees that can be generated from $[1, i]$. Initially, $f[0] = 1$, and the answer is $f[n]$.
We can enumerate the number of nodes $i$, then the number of nodes in the left subtree $j \in [0, i - 1]$, and the number of nodes in the right subtree $k = i - j - 1$. The number of combinations of the number of nodes in the left subtree and the right subtree is $f[j] \times f[k]$, so $f[i] = \sum_{j = 0}^{i - 1} f[j] \times f[i - j - 1]$.
Finally, return $f[n]$.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes.
class Solution:
def numTrees(self, n: int) -> int:
f = [1] + [0] * n
for i in range(n + 1):
for j in range(i):
f[i] += f[j] * f[i - j - 1]
return f[n]
class Solution {
public int numTrees(int n) {
int[] f = new int[n + 1];
f[0] = 1;
for (int i = 1; i <= n; ++i) {
for (int j = 0; j < i; ++j) {
f[i] += f[j] * f[i - j - 1];
}
}
return f[n];
}
}
class Solution {
public:
int numTrees(int n) {
vector<int> f(n + 1);
f[0] = 1;
for (int i = 1; i <= n; ++i) {
for (int j = 0; j < i; ++j) {
f[i] += f[j] * f[i - j - 1];
}
}
return f[n];
}
};
func numTrees(n int) int {
f := make([]int, n+1)
f[0] = 1
for i := 1; i <= n; i++ {
for j := 0; j < i; j++ {
f[i] += f[j] * f[i-j-1]
}
}
return f[n]
}
function numTrees(n: number): number {
const f: number[] = Array(n + 1).fill(0);
f[0] = 1;
for (let i = 1; i <= n; ++i) {
for (let j = 0; j < i; ++j) {
f[i] += f[j] * f[i - j - 1];
}
}
return f[n];
}
impl Solution {
pub fn num_trees(n: i32) -> i32 {
let n = n as usize;
let mut f = vec![0; n + 1];
f[0] = 1;
for i in 1..=n {
for j in 0..i {
f[i] += f[j] * f[i - j - 1];
}
}
f[n] as i32
}
}
public class Solution {
public int NumTrees(int n) {
int[] f = new int[n + 1];
f[0] = 1;
for (int i = 1; i <= n; ++i) {
for (int j = 0; j < i; ++j) {
f[i] += f[j] * f[i - j - 1];
}
}
return f[n];
}
}
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