# 095. Unique Binary Search Trees II

Difficulty: Medium

Frequency: N/A

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.

Test Cases:

Solution 1:

Data structure:
Steps:
Complexity:
Runtime:
Space:
Code:
/**
* Definition for a binary tree node.
* 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) return vector<TreeNode*>();
return genBST(1, n);
}
vector<TreeNode*> genBST(int min, int max) {
vector<TreeNode*> results;
if (min > max) {
results.push_back(NULL);
return results;
}
for (int i = min; i <= max; i++) {
vector<TreeNode*> left = genBST(min, i - 1);
vector<TreeNode*> right = genBST(i + 1, max);
for (int j = 0; j < left.size(); j++) {
for (int k = 0; k < right.size(); k++) {
TreeNode *root = new TreeNode(i);
root->left = left[j];
root->right = right[k];
results.push_back(root);
}
}
}
return results;
}
};

Solution 2:

Data structure:
steps:
Complexity:
Runtime:
Space:
Code:

Submission errors:

Things to learn:

# 096. Unique Binary Search Trees

Difficulty: Medium

Frequency: N/A

Given n, how many structurally unique BST’s (binary search trees) that store values 1…n?

For example,
Given n = 3, there are a total of 5 unique BST’s.

1         3     3      2      1
\       /     /      / \      \
3     2     1      1   3      2
/     /       \                 \
2     1         2                 3

Test Cases:

Solution 1:

Data structure:
BST
Steps:
Complexity:
Runtime:
Space:
Code:
class Solution {
public:
int numTrees(int n) {
vector<int> types(n + 1, 0);
types[0] = types[1] = 1;
for (int i = 2; i <= n; i++) {
for (int j = 1; j <= i; j++) {
types[i] += types[j - 1] * types[i - j];
}
}
return types[n];
}
};

Solution 2:

Data structure:
steps:
Complexity:
Runtime:
Space:
Code:

Submission errors:

Things to learn: