95. Unique Binary Search Trees II

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;
}