代码随想录算法训练营第23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

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

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

538.把二叉搜索树转换为累加树

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