Description
Given an integer n, return all the structurally unique **BST’**s (binary search trees), which has exactly n nodes of unique values from 1 to n. Return the answer in any order.
Example 1:
Input: n = 3 Output: [[1,null,2,null,3],[1,null,3,2],[2,1,3],[3,1,null,null,2],[3,2,null,1]]
Example 2:
Input: n = 1 Output: 1
Constraints:
1 <= n <= 8
Code
time complexity 就是 Catalan Numbers。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<TreeNode*> generateTrees(int n) {
return buildTree(1, n);
}
vector<TreeNode*> buildTree(int start, int end) {
vector<TreeNode*> ans;
// If start > end, then subtree will be empty so add NULL in the ans and return it.
if(start > end) {
ans.push_back(NULL);
return ans;
}
// Iterate through all values from start to end to construct left and right subtree recursively
for(int i = start; i <= end; ++i) {
vector<TreeNode*> leftSubTree = buildTree(start, i - 1); // Construct left subtree
vector<TreeNode*> rightSubTree = buildTree(i + 1, end); // Construct right subtree
// loop through all left and right subtrees and connect them to ith root
for(int j = 0; j < leftSubTree.size(); j++) {
for(int k = 0; k < rightSubTree.size(); k++) {
TreeNode* root = new TreeNode(i); // Create root with value i
root->left = leftSubTree[j]; // Connect left subtree rooted at leftSubTree[j]
root->right = rightSubTree[k]; // Connect right subtree rooted at rightSubTree[k]
ans.push_back(root); // Add this tree(rooted at i) to ans data-structure
}
}
}
return ans;
}
};