Leetcode-95.Unique Binary Search Trees II(a representative recursion problem)

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题目：

Given an integer 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

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AC代码：

class Solution { public: vector<TreeNode*> generateTrees(int n) { vector<TreeNode *>result; if (n == 0)return result; result = generateSubTree(1,n); return result; } vector<TreeNode *> generateSubTree(int start,int end){//产生从start到end之间编号的子树 vector<TreeNode *>re; if (end < start)return re; if (end - start == 0){ TreeNode *temp = new TreeNode(end); re.push_back(temp); return re; } for (int root = start; root <= end;root++){//根节点编号 vector<TreeNode *>left = generateSubTree(start,root-1);//产生左子树 vector<TreeNode *>right = generateSubTree(root+1,end);//产生右子树 int leftnum = left.size(); int rightnum = right.size(); for (int i = 0; i < (leftnum==0?1:leftnum);i++){ for (int j = 0; j < (rightnum==0?1:rightnum);j++){ TreeNode *tem = new TreeNode(root); tem->left = (leftnum==0?NULL:left[i]); tem->right = (rightnum==0?NULL:right[j]); re.push_back(tem); } } } return re; } };

分析：这道题是Leetcode-95.Unique Binary Search Trees的升级，基本思想是类似的，但是这里需要注意的是vector<TreeNode*>存放的是每棵树的根节点

的指针，容器的大小代表其存放的二叉搜索树的数量。在递归求解中，

1.将每个节点的指针依次存放到根节点的左右子树区域(left和right)。

2.通过数字root来划分左右字数的数字，并将它们形成的子二叉搜索树链接起来。。。

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