Unique Binary Search Trees II

Given n, generate all structurally unique BST's (binary search trees) that store values 1...n.

For example,

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

Analysis: DFS. Select different element in the in-order array as the root of each subtree. 
/**
* 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> helper(int[] inorder) {
ArrayList<TreeNode> res = new ArrayList<TreeNode>();
if(inorder.length == 0) {
TreeNode root = null;
res.add(root);
return res;
}

for(int i=0; i<inorder.length; i++) {
ArrayList<TreeNode> left = helper(Arrays.copyOfRange(inorder, 0, i));
ArrayList<TreeNode> right = helper(Arrays.copyOfRange(inorder, i+1, inorder.length));

for(int j=0; j<left.size(); j++) {
for(int k=0; k<right.size(); k++) {
TreeNode root = new TreeNode(inorder[i]);
root.left = left.get(j);
root.right = right.get(k);
res.add(root);
}
}
}
return res;
}

public ArrayList<TreeNode> generateTrees(int n) {
int[] inorder = new int
;
for(int i=0; i<n; i++) {
inorder[i] = i+1;
}
return helper(inorder);
}
}