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/

**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]

**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.

**Solution:**

```
class Solution {
public TreeNode convertBST(TreeNode root) {
Stack<TreeNode> stack = new Stack<>();
TreeNode node = root;
int sum=0;
while(!stack.isEmpty() || node!=null)
{
while(node!=null)
{
stack.add(node);
node=node.right;
}
node = stack.pop();
sum+=node.val;
node.val=sum;
node = node.left;
}
return root;
}
}
```

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