Given an integer n, generate all structurally unique BST’s (binary search trees) that store values 1 … n.

Example:

Input: 3
Output:
[
  [1,null,3,2],
  [3,2,null,1],
  [3,1,null,null,2],
  [2,1,3],
  [1,null,2,null,3]
]
Explanation:
The above output corresponds to the 5 unique BST's shown below:

   1         3     3      2      1
    \       /     /      / \      \
     3     2     1      1   3      2
    /     /       \                 \
   2     1         2                 3

难度：medium

题目：给出整数n，生成其所有可能BST.

思路：递归，分别把1到n作为根结点，例如i, 则1到i-1作为左子树，i+1到n作为右子树。

Runtime: 2 ms, faster than 89.21% of Java online submissions for Unique Binary Search Trees II.
Memory Usage: 39.9 MB, less than 100.00% of Java online submissions for Unique Binary Search Trees II.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<TreeNode> generateTrees(int n) {
        if (n <= 0) {
            return new ArrayList<TreeNode>();
        }
        
        return generateTrees(1, n);
    }
    
    private List<TreeNode> generateTrees(int left, int right) {
        if (left > right) {
            List<TreeNode> treeNodes = new ArrayList<>();
            treeNodes.add(null);
            return treeNodes;
        }
        
        List<TreeNode> result = new ArrayList<>();
        for (int i = left; i <= right; i++) {
            List<TreeNode> leftList = generateTrees(left, i - 1);
            List<TreeNode> rightList = generateTrees(i + 1, right);
            
            for (TreeNode lChild: leftList) {
                for (TreeNode rChild: rightList) {
                    TreeNode root = new TreeNode(i);
                    root.left = lChild;
                    root.right = rChild;
                    result.add(root);
                }
            }
        }
        return result;
    }
}