538. Convert BST to Greater Tree ----Medium

Given the root of a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus the sum of all keys greater than the original key in BST.

As a reminder, a binary search tree is a tree that satisfies these constraints:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than the node's key.
• Both the left and right subtrees must also be binary search trees.

 

Example 1:

Input: root = [4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
Output: [30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]

Example 2:

Input: root = [0,null,1]
Output: [1,null,1]

 

Constraints:

• The number of nodes in the tree is in the range [0, 104].
• -104 <= Node.val <= 104
• All the values in the tree are unique.
• root is guaranteed to be a valid binary search tree.

A:

-------------利用  BST -----------

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* convertBST(TreeNode* root) {
        int total = 0;
        helper(root, total);
        return root;
    }
private:
    void helper(TreeNode* root, int & total){
        if(!root)
            return;
        helper(root->right, total);
        root->val += total;
        total = root->val;
        helper(root->left, total);
    }
};

Mistakes：
   root->val  += S
  S+= root->val      这两句是错误的， 需要利用tmp来保存临时值