# 0538. Convert BST to Greater Tree

https://leetcode.com/problems/convert-bst-to-greater-tree

## Description

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 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.

**Note:** This question is the same as 1038: [https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/](https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/)

**Example 1:**

![](https://assets.leetcode.com/uploads/2019/05/02/tree.png)

```
**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]
```

**Example 3:**

```
**Input:** root = [1,0,2]
**Output:** [3,3,2]
```

**Example 4:**

```
**Input:** root = [3,2,4,1]
**Output:** [7,9,4,10]
```

**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.

## ac

```java
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int sum = 0;

    public TreeNode convertBST(TreeNode root) {
        if (root == null) return root;
        convertBST(root.right);
        sum += root.val;
        root.val = sum;
        convertBST(root.left);
        return root;
    }
}

/*
1) get total sum; 2) in-order traversal, total sum - prev sum;
update from right to left, add each node.val to sum, update node.val = sum
*/
```
