Given an integer n, generate all structurally unique BST‘s (binary search trees) that store values 1 ... n.
Example:
Input: 3
Output:
[
[1,null,3,2],
[3,2,null,1],
[3,1,null,null,2],
[2,1,3],
[1,null,2,null,3]
]
Explanation:
The above output corresponds to the 5 unique BST‘s shown below:
1 3 3 2 1
\ / / / \ 3 2 1 1 3 2
/ / \ 2 1 2 3
ATTENTION ＰＯＩＮＴ
defination of tree
#include <iostream>
#include<vector>
using namespace std;
#include <iostream>
#include<vector>
//segement default内存操作不当
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 {};
const auto& ans=generateTrees(1,n);
cout<<ans.size()<<endl;
return ans;
}
private:
vector<TreeNode*> generateTrees(int l,int r)
{
if(l>r) return {nullptr};
vector<TreeNode*> ans;
for(int i=l;i<=r;++i)
for(TreeNode* left:generateTrees(l,i-1))
for(TreeNode* right:generateTrees(i+1,r))
{
ans.push_back(new TreeNode(i));
ans.back()->left=left;
ans.back()->right=right;
}
return ans;
}
};
//class Solution {
//public:
// vector<TreeNode*> generateTrees(int n) {
// if (n == 0) return {};
// const auto& ans = generateTrees(1, n);
// cout << ans.size() << endl;
// return ans;
// }
//private:
// vector<TreeNode*> generateTrees(int l, int r) {
// if (l > r) return { nullptr };
// vector<TreeNode*> ans;
// for (int i = l; i <= r; ++i)
// for (TreeNode* left : generateTrees(l, i - 1))
// for (TreeNode* right : generateTrees(i + 1, r)) {
// ans.push_back(new TreeNode(i));
// ans.back()->left = left;
// ans.back()->right = right;
// }
// return ans;
// }
//};
void printTree(TreeNode* tmp)
{
if(tmp==nullptr)
return;
else{
cout<<tmp->val;
printTree(tmp->left);
printTree(tmp->right);
}
}
int main()
{
Solution s1;
int n=3;
// cin>>n;
vector<TreeNode*> ans=s1.generateTrees(n);
for(auto i:ans)
{
printTree(i);
cout<<endl;
}
return 0;
}
Leetcode开心刷题三十九——95. Unique Binary Search Trees II
原文：https://www.cnblogs.com/Marigolci/p/11516205.html