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

This is the second time I solve this problem.

Everytime when we encounter a BST problem without quite clear thought, we can resort to Divide & Conquer.

Use recursion to build the left and right subtree, then combine them then return.

In this problem, the left and right subtree can be multiple.

FIRST TRY ERROR: Forget to clear the vector of left of right subtree while apply different root.

Code:

/**
* 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 *> res;
if(n == 0)
{
res.push_back(NULL);
return res;
} res = gen(1, n);
return res;
} vector<TreeNode *> gen(int start, int end)
{
vector<TreeNode*> res;
if(start == end)
{
TreeNode* tmp = new TreeNode(start);
res.push_back(tmp);
return res;
} vector<TreeNode*> leftsub, rightsub;
for(int i = start; i <= end; i++)
{
leftsub.clear(); // First try error
rightsub.clear(); // First try error if(i == start) leftsub.push_back(NULL);
else leftsub = gen(start, i-1); if(i == end) rightsub.push_back(NULL);
else rightsub = gen(i+1, end); for(int m = 0; m < leftsub.size(); m++)
{
for(int n = 0; n < rightsub.size(); n++)
{
TreeNode* root = new TreeNode(i); // divide & conquer
root->left = leftsub[m];
root->right = rightsub[n];
res.push_back(root);
}
}
} return res;
}
};

  

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