[LeetCode] 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.

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思路：Unique Binary Search Trees是用dp，求解所有的trees不能用dp。

对于每一个root，有leftNum * rightNum种情况，直接穷举，对于start>end, 返回含有NULL的一个节点，start==end也返回一个节点，框架和深搜一样

class Solution {
    public:
        vector<TreeNode *> dfs(int start, int end)
        {
            vector<TreeNode*> trees;

            if(start > end)
            {
                trees.push_back(NULL);
                return trees;
            }
            if(start == end)
            {
                TreeNode* p = new TreeNode(start);
                trees.push_back(p);
                return trees;
            }

            for(int rootVal = start; rootVal <= end; rootVal ++)
            {
                vector<TreeNode*> leftSubTrees = dfs(start, rootVal - 1);
                vector<TreeNode*> rightSubTrees = dfs(rootVal + 1, end);

                int leftSubTreesNum = leftSubTrees.size();
                int rightSubTreesNum = rightSubTrees.size();
                for(int i = 0; i < leftSubTreesNum; i++)
                {
                    for(int j= 0; j< rightSubTreesNum; j++)
                    {
                        TreeNode* root = new TreeNode(rootVal);
                        TreeNode* left = leftSubTrees[i];
                        TreeNode* right= rightSubTrees[j];
                        root->left = left;
                        root->right= right;
                        trees.push_back(root);
                    }
                }
            }
            return trees;
        }
        vector<TreeNode *> generateTrees(int n) {
            if(n == 0)
            {
                vector<TreeNode *>   trees;
                trees.push_back(NULL);
                return trees;
            }
            return dfs(1, n);
        }
};