代码随想录day21

669.修剪二叉搜索树

//理解修建后重建树的概念

    TreeNode* trimBST(TreeNode* root, int low, int high) {
        if(root == nullptr) return nullptr;
        if(root->val < low){
            TreeNode* node = trimBST(root->right, low, high);
            return node;
        }
        if(root->val > high){
            TreeNode* node = trimBST(root->left, low, high);
            return node;
        }
        root->left = trimBST(root->left, low, high);
        root->right = trimBST(root->right, low, high);
        return root;
    }

108.将有序数组转化为二叉搜索树

    TreeNode* sortedArrayToBST(vector<int>& nums) {
        int sz = nums.size();
        if(sz == 0) return nullptr;
        if(sz == 1) return new TreeNode(nums[0]);
        int mid = sz / 2;
        TreeNode* root = new TreeNode(nums[mid]);
        vector<int> left(nums.begin(), nums.begin()+mid);
        vector<int> right(nums.begin()+mid+1, nums.end());
        root->left = sortedArrayToBST(left);
        root->right = sortedArrayToBST(right);
        return root;
        
    }

538.把二叉搜索树转换为累加树

//注意双指针法的变通，以及右中左遍历

    int prev;
    void traverse(TreeNode* root){
        if(root == nullptr) return;
        
        traverse(root->right);
        root->val += prev;
        prev = root->val;
        traverse(root->left);
    }
    TreeNode* convertBST(TreeNode* root) {
        prev = 0;
        traverse(root);
        return root;
    }