LeetCode 95. Unique Binary Search Trees II
Given an integer 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
分析
这道题的思路一开始想错了,参考了discussion,当n=m时的结果很容易想到应该是在n=m-1时的结果上得到的,相当于在n=m-1的各棵树插入n这个节点,分为插在根节点上和右子树上两种情况。
/**
* 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:
TreeNode* clone(TreeNode* root){
if(root == nullptr)
return nullptr;
TreeNode* newroot = new TreeNode(root->val);
newroot->left = clone(root->left);
newroot->right = clone(root->right);
return newroot;
}
vector<TreeNode *> generateTrees(int n) {
vector<TreeNode*> r;
if(n==0) return r;
vector<TreeNode *> res(1,nullptr);
for(int i = 1; i <= n; i++){
vector<TreeNode *> tmp;
for(int j = 0; j<res.size();j++){
TreeNode* oldroot = res[j];
TreeNode* root = new TreeNode(i);
TreeNode* target = clone(oldroot);
root->left = target;
tmp.push_back(root);
if(oldroot!=nullptr){
TreeNode* tmpold = oldroot;
while(tmpold->right!=nullptr){
TreeNode* nonroot = new TreeNode(i);
TreeNode *tright = tmpold->right;
tmpold->right = nonroot;
nonroot->left = tright;
TreeNode *target = clone(oldroot);
tmp.push_back(target);
tmpold->right = tright;
tmpold = tmpold->right;
}
tmpold->right = new TreeNode(i);
TreeNode *target = clone(oldroot);
tmp.push_back(target);
tmpold->right = nullptr;
}
}
res=tmp;
}
return res;
}
};