add py3 soln

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

@@ -0,0 +1,49 @@
+# 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