add py3 soln

0538_convert-bst-to-greater-tree/README.md

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+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:**
+
+![](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]
+    
+
+**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.
+
+**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/)

0538_convert-bst-to-greater-tree/python3/solution.py

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+# 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:
+    # Perform reverse in-order traversal of the given BST.
+    # This will give us numbers in descending order as we
+    # touch each node.
+    #
+    # Right subtree -> Current node -> Left subtree
+    #
+    def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
+        self.walk(root, 0)
+        
+        return root
+    
+    # `total` will be the state used to keep the sum of
+    # all numbers greater than `node.val`
+    def walk(self, node, total):
+        if node is not None:
+            # Since all nums to the right will be greater
+            # than current, we need to get the total of
+            # the right subtree
+            right_total = self.walk(node.right, total)
+            
+            # Need to store the new total based on the right
+            # subtree's total we obtained
+            next_total = node.val + right_total
+            
+            # Update current node's total to its value plus
+            # total of all numbers greater than it (i.e right
+            # subtree)
+            node.val = next_total
+            
+            # If there's a left subtree, it could mean that the
+            # there might be a new total that we need to return one
+            # level down the call stack
+            next_total = self.walk(node.left, next_total)
+            
+            return next_total
+        
+        # If node isn't present, we can't just return 0.
+        #
+        # 0 would be valid for the very first rightmost subtree.
+        # But as soon as we touch left subtrees, `total` could
+        # be different from initial 0 we pass
+        return total