标签：
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.
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-.-!!!
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<TreeNode *> generateTrees(int n) { 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; } };
[LeetCode] Unique Binary Search Trees II 独一无二的二叉搜索树之二
标签：
原文地址：http://www.cnblogs.com/grandyang/p/4301096.html
