95. Unique Binary Search Trees II
Đề Bài
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
Thuật Toán & Kỹ Thuật
⏱️ Thời gian
O(V+E)
💾 Không gian
O(V)
Lời Giải
C++
0095-unique-binary-search-trees-ii.cpp
/**
* 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*> dfs(int s, int e) {
if (s > e) {
return {nullptr};
}
vector<TreeNode*> res;
for (int i = s; i <= e; ++i) {
vector<TreeNode*> left_node = dfs(s, i - 1);
vector<TreeNode*> right_node = dfs(i + 1, e);
for (auto a : left_node) {
for (auto b : right_node) {
TreeNode* root = new TreeNode(i);
root->left = a;
root->right = b;
res.push_back(root);
}
}
}
return res;
}
vector<TreeNode*> generateTrees(int n) {
return dfs(1, n);
}
};