Unique Binary Search Trees

Given N, how many structurally uniqueBST'S (Binary Search Trees) that store values 1... n?

For example,

GivenN= 3, there are a total of 5 unique BST's.

   1         3     3      2      1    \       /     /      / \           3     2     1      1   3      2    /     /       \                    2     1         2                 3

Idea: If I is taken as the vertex, the corresponding number of Binary Trees is: numtrees (I-1) * numtrees (n-I ). You can solve this problem by adding different I. Here, to facilitate recursion, set numtrees (0) to 1. The recursive solution is as follows:

 1 class Solution { 2 public: 3     int numTrees( int n ) { 4         if( n == 0 ) { return 1; } 5         if( n <= 2 ) { return n; } 6         int num = 0; 7         for( int i = 1; i <= n/2; ++i ) { 8             num += 2 * numTrees( i-1 ) * numTrees( n-i ); 9         }10         if( n % 2 == 1 ) { num += numTrees( n/2 ) * numTrees( n/2 ); }11         return num;12     }13 };

Use num [I] To save numtrees (I). The dynamic planning solution is as follows:

 1 class Solution { 2 public: 3     int numTrees(int n) { 4         if( n <= 2 ) { return n; } 5         vector<int> num( n, 0 ); 6         num[0] = 1; num[1] = 2; 7         for( int i = 2; i < n; ++i ) { 8             num[i] += 2*num[i-1]; 9             for( int j = 1; j < i; ++j ) {10                 num[i] += num[j-1] * num[i-j-1];11             }12         }13         return num.back();14     }15 };

The results meet the catlan recursive formula: H (n) = H (0) * H (n-1) + H (1) * H (n-2) + H (2) * H (n-3) +... + H (I) * H (n-i-1) +... + H (n-1) * H (0) (N> = 2), and H (0) = 1, H (1) = 1. Another recursive form is H (n) = H (n-1) * (4 * N-2)/(n + 1 ).

 1 class Solution { 2 public: 3     int numTrees( int n ) { 4         if( n <= 1 ) { return n; } 5         int catalan = 1; 6         for( int i = 2; i <= n; ++i ) { 7             catalan = catalan * (4*i-2) / (i+1); 8         } 9         return catalan;10     }11 };