代码随想录算法训练营第19天| 669.修剪二叉搜索树、108.将有序数组转换为二叉搜索树、538.把二叉搜索树转换为累加树

669. 修剪二叉搜索树

题目链接

提交代码

class Solution {
public:TreeNode* trimBST(TreeNode* root, int low, int high) {if(root == nullptr) return nullptr;if(root -> val < low){root -> right = trimBST(root -> right, low, high);return root -> right;}else if(root ->val > high){root ->left = trimBST(root -> left, low, high);return root -> left;}  root ->left = trimBST(root -> left, low, high);root -> right = trimBST(root -> right, low, high);return root;}
};

108.将有序数组转换为二叉搜索树

题目链接

提交代码（方法）

class Solution {
private:TreeNode* traversal(vector<int>& nums, int left, int right) {if (left > right) return nullptr;int mid = left + ((right - left) / 2);TreeNode* root = new TreeNode(nums[mid]);root->left = traversal(nums, left, mid - 1);root->right = traversal(nums, mid + 1, right);return root;}
public:TreeNode* sortedArrayToBST(vector<int>& nums) {TreeNode* root = traversal(nums, 0, nums.size() - 1);return root;}
};

538.把二叉搜索树转换为累加树

题目链接

提交代码（方法）

class Solution {
public:int cur = 0;TreeNode* convertBST(TreeNode* root) {if(root == nullptr) return nullptr;convertBST(root -> right);root -> val = cur + root -> val;cur = root -> val;convertBST(root -> left);return root;}
};

总结

                     日期： 2023 年 4 月 4 日
              学习时长： 0 h 30 m
                     难度：$★$
累计完成题目数量： 60
距离目标还有数量： 240
                      小结：