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

Note:
可行的二叉查找树的数量是相应的卡特兰数.
算法上还是用求解NP问题的方法来求解，也就是N-Queens中介绍的在循环中调用递归函数求解子问题。思路是每次一次选取一个结点为根，然后递归求解左右子树的所有结果，最后根据左右子树的返回的所有子树，依次选取然后接上（每个左边的子树跟所有右边的子树匹配，而每个右边的子树也要跟所有的左边子树匹配，总共有左右子树数量的乘积种情况），构造好之后作为当前树的结果返回。

/**
 * 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 generateTrees(int n) {
        return helper(1, n);
    }
    
    public ArrayList helper(int start, int end) {
        ArrayList res = new ArrayList();
        
        if(start > end) {
            res.add(null);
            return res;
        }
        
        for(int i = start; i <= end; i++) {
            ArrayList leftTree = helper(start, i - 1);
            ArrayList rightTree = helper(i + 1, end);
            for(int j = 0; j < leftTree.size(); j++) {
                for(int k = 0; k < rightTree.size(); k++) {
                    TreeNode root = new TreeNode(i);
                    root.left = leftTree.get(j);
                    root.right = rightTree.get(k);
                    res.add(root);
                }
            }
        }
        return res;
    }
}