LC.95.不同的二叉搜索树 II(递归)
思路:递归实现,考虑枚举根结点,递归左右子树,然后从左子树的答案和右子树的答案选一个合并丢进 里即可。
/**
* 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 l,int r){
if(l>r) return {nullptr};
vector<TreeNode*> ans;
for(int i=l;i<=r;i++){
vector<TreeNode*> left_tree=dfs(l,i-1);
vector<TreeNode*> right_tree=dfs(i+1,r);
for(auto j:left_tree)
for(auto k:right_tree){
TreeNode * Cur=new TreeNode(i);
Cur->left=j;
Cur->right=k;
ans.push_back(Cur);
}
}
return ans;
}
vector<TreeNode*> generateTrees(int n) {
if(!n) return {};
return dfs(1,n);
}
};