Unique Binary Search Trees

当数组为 1，2，3，4，.. i，.. n时，基于以下原则的BST建树具有唯一性：

以i为根节点的树，其左子树由[0, i-1]构成， 其右子树由[i+1, n]构成。

再由递归便可得:

public static int numTrees(int n) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int[] myarray = new int[n+1];
        return getNumTree(n,myarray);
    }


public static int getNumTree(int n, int[] myarray){
        if(myarray[n] != 0)
            return myarray[n];
        if(n == 0)
        {
            myarray[0] = 1;
            return 1;
        }
        if(n == 1)
        {
            myarray[1] = 1;
            return 1;
        }
        int count = 0;
        for(int i = 0; i < n; i++)
            {
                count += getNumTree(i,myarray) * getNumTree(n-1-i,myarray);
            }
        myarray[n] = count;
        return myarray[n];
    }