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

方法：遇到這種問題不能用深度優先，要用自頂向下，分而治之。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private List<TreeNode> generate(int from, int to) {
        List<TreeNode> roots = new ArrayList<>();
        if (from == to) {
            roots.add(new TreeNode(from));
            return roots;
        }
        for(int i=from; i<=to; i++) {
            if (i==from) {
                List<TreeNode> rights = generate(from+1, to);
                for(int j=0; j<rights.size(); j++) {
                    TreeNode root = new TreeNode(from);
                    root.right = rights.get(j);
                    roots.add(root);
                }
            } else if (i==to) {
                List<TreeNode> lefts = generate(from, to-1);
                for(int j=0; j<lefts.size(); j++) {
                    TreeNode root = new TreeNode(to);
                    root.left = lefts.get(j);
                    roots.add(root);
                }
            } else {
                List<TreeNode> lefts = generate(from, i-1);
                List<TreeNode> rights = generate(i+1, to);
                for(int j=0; j<lefts.size(); j++) {
                    for(int k=0; k<rights.size(); k++) {
                        TreeNode root = new TreeNode(i);
                        root.left = lefts.get(j);
                        root.right = rights.get(k);
                        roots.add(root);
                    }
                }
            }
        }
        return roots;
    }
    public List<TreeNode> generateTrees(int n) {
        return generate(1, n);
    }
}