95. Unique Binary Search Trees II （独一无二的二叉搜索树之二）

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

confused what “{1,#,2,3}” means? > read more on how binary tree is serialized on OJ.

OJ’s Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where ‘#’ signifies a path terminator where no node exists below.

Here’s an examp
1
/ \
2 3
/
4
\
5
above binary tree is serialized as “{1,2,3,#,#,4,#,#,5}”.

这道题是之前的 Unique Binary Search Trees 独一无二的二叉搜索树的延伸，之前那个只要求算出所有不同的二叉搜索树的个数，这道题让把那些二叉树都建立出来。这种建树问题一般来说都是用递归来解，这道题也不例外，划分左右子树，递归构造。至于递归函数中为啥都用的是指针，是参考了网友水中的鱼的博客，若不用指针，全部实例化的话会存在大量的对象拷贝，要调用拷贝构造函数，具体我也不太懂，反正感觉挺有道理的，不明觉厉啊-.-!!!

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;    }};

笔记：这种递归调用涉及到了指针和vector