LeetCode OJ——Unique Binary Search Trees II

http://oj.leetcode.com/problems/unique-binary-search-trees-ii/

一题要求得出所有树的种类数，二题要求得出所有树。

在一题的基础上修改代码，还是要提前想清楚再写。

#include <iostream>
#include <vector>
using namespace std;

struct TreeNode {
         int val;
         TreeNode *left;
         TreeNode *right;
         TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     };

class Solution {
public:
    vector<TreeNode *> fun(int start,int end )
    {
        vector<TreeNode *> ansTemp;
        vector<TreeNode *> ansTemp2;
        vector<TreeNode *> ansTemp3;

        ansTemp.clear();
        if(start>end)
        {
            ansTemp.push_back(NULL);
            return ansTemp;
        }
        if(start == end)
        {
            TreeNode *mid = new TreeNode(start);
            ansTemp.push_back(mid);
            return ansTemp;
        }
    
        for(int i=start;i<=end;i++)//枚举的中间位置
        {
            ansTemp2 = fun(start,i-1);
            ansTemp3 = fun(i+1,end);
            
            if(ansTemp2.size()!=0&&ansTemp3.size()!=0)
                for(int j = 0;j<ansTemp2.size();j++)
                {
                    for(int k = 0;k<ansTemp3.size();k++)
                    {
                        TreeNode * root = new TreeNode(0);
                        root->val = i;
                        root->left = ansTemp2[j];
                        root->right = ansTemp3[k];
                        ansTemp.push_back(root);
                    }
                }
            else if(ansTemp2.size()!=0&&ansTemp3.size()==0)
                for(int j = 0;j<ansTemp2.size();j++)
                {
                    TreeNode * root = new TreeNode(0);
                    root->val = i;
                    root->left = ansTemp2[j];
                    root->right = NULL;
                    ansTemp.push_back(root);
                }
            else if(ansTemp2.size()==0&&ansTemp3.size()!=0)
                for(int k = 0;k<ansTemp3.size();k++)
                {
                    TreeNode * root = new TreeNode(0);
                    root->val = i;
                    root->left = NULL;
                    root->right = ansTemp3[k];
                    ansTemp.push_back(root);
                }
            else
            {
                TreeNode * root = new TreeNode(i);
                ansTemp.push_back(root);
            }

        }
        
        return ansTemp;
    }

    vector<TreeNode *> generateTrees(int n) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        vector<TreeNode *> ans;
        ans.clear();        
        ans =  fun(1,n);
        return ans;
    }
};
 
int main()
{
    Solution my;
    my.generateTrees(0);
    return 0;
}

亲爱的，加油。