95. Unique Binary Search Trees II

Given an integer 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

牵扯到列举，用递归求解。设置start和end，遍历i为root，产生i左右两边的list<node>，left和right，再组合left，i，和right，返回结果。代码如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<TreeNode> generateTrees(int n) {
        if (n <= 0) {
            return new ArrayList<TreeNode>();
        }
        return helper(1, n);
    }
    
    private List<TreeNode> helper(int start, int end) {
        List<TreeNode> list = new ArrayList<TreeNode>();
        if (start > end) {
            list.add(null);
            return list;
        }
        if (start == end) {
            list.add(new TreeNode(start));
            return list;
        }
        List<TreeNode> left, right;
        for(int i = start; i <= end; i ++) {
            left = helper(start, i - 1);
            right = helper(i + 1, end);
            for (TreeNode nleft: left) {
                for (TreeNode nright: right) {
                    TreeNode root = new TreeNode(i);
                    root.left = nleft;
                    root.right = nright;
                    list.add(root);
                }
            }
        }
        return list;
    }
}