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 example:
1
/ \
2 3
/
4
\
5
The 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; } };
本文转自博客园Grandyang的博客，原文链接：独一无二的二叉搜索树之二[LeetCode] Unique Binary Search Trees II ，如需转载请自行联系原博主。
