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/**
 * Given n, generate all structurally unique BST's (binary search trees) that store values 1...n.
 * Given n = 3, your program should return all 5 unique BST's shown below.
 * 1         3     3      2      1
    \       /     /      / \      \
     3     2     1      1   3      2
    /     /       \                 \
   2     1         2                 3
 */
 
/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; left = null; right = null; }
 * }
 */ 

public class Solution{
    public ArrayList<TreeNode> generateTrees(int n) {
        if(n == 0) return generateTrees(1, 0);
        return generateTrees(1, n);
        
    }
    public ArrayList<TreeNode> generateTrees(int start, int end) {
        ArrayList<TreeNode> subTrees = new ArrayList<TreeNode>();
        if(start > end){
            subTrees.add(null);
            return subTrees;
        }          
        for(int i = start; i <= end; i++){
            for(TreeNode left : generateTree(start, i - 1)){
                for(TreeNode right : generateTrees(i + 1, end)){
                    TreeNode aTree = new TreeNode(i);
                    aTree.left = left;
                    aTree.right = right;
                    subTrees.add(aTree);
                }
            }
        }
        return subTrees;
    }    
}

/**
 * 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
*/


public class Solution{
    public int numTrees(int n) {
        int[] count = new int[n + 1];
        count[0] = 1;
        count[1] = 1;
        for(int i = 2; i <= n; i++){
            for(int j = 0; j < i; j++){
                count[i] += count[j] * count[i - j - 1];
            }
        }
        return count[n];
    }
    
}