代码随想录算法训练营43期 | Day

108.将有序数组转换为二叉搜索树

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

class Solution {
public:
    int pre = 0;
    void traversal(TreeNode* cur)
    {
        // 终止条件
        if(cur==nullptr) return;
        //右
        traversal(cur->right);
        cur->val += pre;
        pre = cur->val;
        traversal(cur->left);
    }

    TreeNode* convertBST(TreeNode* root) {
        pre = 0;
        traversal(root);
        return root;
    }
};