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

/**
* 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);
}
}
}
};