LeetCode: 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

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    vector<TreeNode *> generateTrees(int n) {        if(n == 0)            return *buildTree(1, 0);        return *buildTree(1, n);    }private:    vector<TreeNode *>* buildTree(int start, int end)    {        vector<TreeNode* >* cur = new vector<TreeNode *>();        if(start > end)        {            (*cur).push_back(NULL);            return cur;        }        for(int i = start; i <= end; i++)        {            vector<TreeNode* > *left = buildTree(start, i-1);            vector<TreeNode* > *right = buildTree(i+1, end);            for(int l = 0; l < (*left).size(); l++)            {                for(int r = 0; r < (*right).size(); r++)                {                    TreeNode *root = new TreeNode(i);                    root->left = (*left)[l];                    root->right = (*right)[r];                    (*cur).push_back(root);                }            }                    }        return cur;    }};

Round 2:

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    vector<TreeNode *> generateTrees(int n) {        vector<TreeNode *> result;        dfs(1, n, result);        return result;    }private:    void dfs(int start, int end, vector<TreeNode*> &cur)    {        if(start > end)        {            cur.push_back(NULL);            return;        }        for(int i = start; i <= end; i++)        {            vector<TreeNode *> left;            dfs(start, i-1, left);            vector<TreeNode *> right;            dfs(i+1, end, right);            for(int x = 0; x < left.size(); x++)                for(int y = 0; y < right.size(); y++)                {                    TreeNode *node = new TreeNode(i);                    node->left = left[x];                    node->right = right[y];                    cur.push_back(node);                }        }    }};

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