Unique Binary Search Trees in C++

Suppose we have an integer n, we have to count all structurally unique binary search trees that store values from 1 to n. So if the input is 3, then the output will be 5, as the trees will be –

To solve this, we will follow these steps –

• create one array of size n + 1
• dp[0] := 1
• for i := 1 to n
  • for j := 0 to i – 1
    • dp[i] := dp[i] + (dp[i – 1 – j] * dp[j])
• return dp[n]

Example(C++)

Let us see the following implementation to get a better understanding −

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   int numTrees(int n) {
      vector <int> dp(n+1);
      dp[0] = 1;
      for(int i =1;i<=n;i++){
         for(int j = 0;j<i;j++){
            //cout << j << " " << i-1-j << " " << j << endl;
            dp[i] += (dp[i-1-j] * dp[j]);
         }
      }
      return dp[n];
   }
};
main(){
   Solution ob;
   cout << ob.numTrees(4);
}

Input

4

Output

14