95. Unique Binary Search Trees II

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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.

题意：根据数字n生成，由数字1-n对应的所有的二叉搜索树。

思路：采用递归并存储中间结果的思路。即动态规划去解，时间会快很多。

#include<iostream>#include<vector> #include<string>#include<stack>#include<list>#include<stdlib.h>using namespace std;//Definition for a binary tree node.struct TreeNode {     int val;     TreeNode *left;     TreeNode *right;     TreeNode(int x) : val(x), left(NULL), right(NULL) {}};/*听了算法讨论班的课程，课上一同学讲了递归的方法，然后又有人提出了一个存储中间结果的方法，不过存储中间结果的方法同学讲得很乱，遂按照自己的思路写一下存储过程的方法。思路：存储以中间序列点以左的点的树的根 && 存储以中间序列点以右的点的树的根*/class Solution {public:vector<TreeNode*> generateTrees(int n) {if (n == 0){vector<TreeNode* > t;return t;}vector<vector<vector<TreeNode*> > > category;for (int i = 0; i <= n+1; i++){vector<vector<TreeNode*> > ca;for (int j = 0; j <= n+1; j++){vector<TreeNode* > c;ca.push_back(c);}category.push_back(ca);}for (int i = 1; i <= n; i++){category[i][i].push_back(new TreeNode(i));}return travel(1, n, category);}vector<TreeNode*> travel(int start, int end, vector<vector<vector<TreeNode*> > >& category){if (!category[start][end].empty()){return category[start][end];}else{if (start > end){category[start][end].push_back(NULL);return category[start][end];}for (int i = start; i <= end; i++){vector<TreeNode*> a = travel(start, i - 1, category);vector<TreeNode*> b = travel(i + 1, end, category);for (int j = 0; j < a.size(); j++){for (int k = 0; k < b.size(); k++){TreeNode* t = new TreeNode(i);t->left = a[j];t->right = b[k];category[start][end].push_back(t);}}}return category[start][end];}}};int main(){Solution solution;vector<TreeNode*> m;m = solution.generateTrees(0);cout << m.size() << endl;return 0;}

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