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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 <= 19
public int numTrees(int n) {
int [] dp = new int[n+1];
dp[0]= 1;
dp[1] = 1;
for(int level = 2; level <=n; level++)
for(int root = 1; root<=level; root++)
dp[level] += dp[level-root]*dp[root-1];
return dp[n];
}
class Solution {
public:
int dp[20]{};
int numTrees(int n) {
if(n <= 1) return 1;
if(dp[n]) return dp[n];
for(int i = 1; i <= n; i++)
dp[n] += numTrees(i-1) * numTrees(n-i);
return dp[n];
}
};
class Solution:
@cache
def numTrees(self, n: int) -> int:
if n <= 1: return 1
return sum(self.numTrees(i-1) * self.numTrees(n-i) for i in range(1, n+1))
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