Unique Binary Search Trees

Given n, how many structurally unique BST's (binary search trees) that store values 1...n?

For example,
Given n = 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

Let's first see some basic cases,
If n==0, the root is null, then the number of BST is 1. numTrees(0)=1.
If n==1, the root is 1, then the number of BST is 1. numTrees(1)=1.
If n==2, we can either has 1(root)2(right) or 2(root)1(left), then the number of BST is 2. numTrees(2)=2.
The key idea of this problem is that each tree of n can be constructed by setting its root from 1, 2.. to n, and construct it's left tree and right tree recursively. For example, in the above diagram, we iterate from 1 to 3, and for 1, we can only have right subtrees with two nodes, then we calll numTrees(0) *numTrees(2) and add the result. then we pick 2 to be the root, and we can only have 2 and 3 as left and right trees accordingly, which is to call numTrees(1) * numTrees (1). Then we pick 3 as root, we call numTrees(2)*numTrees(0). However, we even do not need a recursion, but use an array to keep intermediate result. Further, you can find that the result is quite symmetric, that we even do not need to iterate a full cycle, but just reach the half, i.e., n/2. The second half should have the same number of trees as the first half.

Java: UniqueBinarySearchTree

01 public int numTrees(int n) {

02     if(n==0)

03         return 1;

04     if(n==1||n==2)

05         return n;

06     int[] array=new int[n+1];

07   

08     array[0]=1;

09     array[1]=1;

10     array[2]=2;

11   

12     for(int i=3;i<=n;i++)

13     {

14         int j;

15         for(j=1;j<=i/2;j++)

16         {

17             array[i]+=array[i-j]*array[j-1]*2;

18         }

19       

20         if(i%2==1)

21         {

22             array[i]+=array[i-j]*array[i-j];

23         }          

24     }      

25     return array[n];      

26 }