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

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

這道題是之前的 Unique Binary Search Trees 獨一無二的二叉搜尋樹的延伸，之前那個只要求算出所有不同的二叉搜尋樹的個數，這道題讓把那些二叉樹都建立出來。這種建樹問題一般來說都是用遞迴來解，這道題也不例外，劃分左右子樹，遞迴構造。至於遞迴函式中為啥都用的是指標，是參考了網友水中的魚的部落格，若不用指標，全部例項化的話會存在大量的物件拷貝，要呼叫拷貝建構函式，具體我也不太懂，反正感覺挺有道理的，不明覺厲啊-.-!!!

class Solution {
public:
    vector<TreeNode *> generateTrees(int
 
 n) {
        if (n == 0) return {};
        return *generateTreesDFS(1, n);
    }
    vector<TreeNode*> *generateTreesDFS(int start, int end) {
        vector<TreeNode*> *subTree = new vector<TreeNode*>();
        if (start > end) subTree->push_back(NULL);
        else {
            for (int i = start; i <= end; ++i) {
                vector<TreeNode*> *leftSubTree = generateTreesDFS(start, i - 1);
                vector<TreeNode*> *rightSubTree = generateTreesDFS(i + 1, end);
                for (int j = 0; j < leftSubTree->size(); ++j) {
                    for (int k = 0; k < rightSubTree->size(); ++k) {
                        TreeNode *node = new TreeNode(i);
                        node->left = (*leftSubTree)[j];
                        node->right = (*rightSubTree)[k];
                        subTree->push_back(node);
                    }
                }
            }
        }
        return subTree;
    }
};