算法训练营补签day25

题目1：669. 修剪二叉搜索树 - 力扣（LeetCode）

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

题目2：108. 将有序数组转换为二叉搜索树 - 力扣（LeetCode）

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

题目3：538. 把二叉搜索树转换为累加树 - 力扣（LeetCode）

class Solution {
public:
    int pre = 0;
    void traversal(TreeNode* node) {
        if(node == NULL) return;
        traversal(node->right);
        node->val = node->val + pre;
        pre = node->val;
        traversal(node->left); 
    }
    TreeNode* convertBST(TreeNode* root) {
        traversal(root);
        return root;
    }
};