Binary Search Tree to Greater Sum Tree
<p>Given the <code>root</code> 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.</p>

<p>As a reminder, a <em>binary search tree</em> is a tree that satisfies these constraints:</p>

<ul>
	<li>The left subtree of a node contains only nodes with keys <strong>less than</strong> the node&#39;s key.</li>
	<li>The right subtree of a node contains only nodes with keys <strong>greater than</strong> the node&#39;s key.</li>
	<li>Both the left and right subtrees must also be binary search trees.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1000-1099/1038.Binary%20Search%20Tree%20to%20Greater%20Sum%20Tree/images/tree.png" style="width: 400px; height: 273px;" />
<pre>
<strong>Input:</strong> root = [4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
<strong>Output:</strong> [30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = [0,null,1]
<strong>Output:</strong> [1,null,1]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 100]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 100</code></li>
	<li>All the values in the tree are <strong>unique</strong>.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Note:</strong> This question is the same as 538: <a href="https://leetcode.com/problems/convert-bst-to-greater-tree/" target="_blank">https://leetcode.com/problems/convert-bst-to-greater-tree/</a></p>
Tree; Depth-First Search; Binary Search Tree; Binary Tree
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def bstToGst(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        def dfs(root: Optional[TreeNode]):
            if root is None:
                return
            dfs(root.right)
            nonlocal s
            s += root.val
            root.val = s
            dfs(root.left)

        s = 0
        dfs(root)
        return root
