Update 0538.把二叉搜索树转换为累加树.md

补充python注释

problems/0538.把二叉搜索树转换为累加树.md

@@ -196,20 +196,40 @@ class Solution {
 ```
 
 ## Python 
+**递归**
 
-递归法
+```python3
-```python
+# 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 convertBST(self, root: TreeNode) -> TreeNode:
+    def __init__(self):
-        def buildalist(root): 
+        self.pre = TreeNode()
-            if not root: return None
+
-            buildalist(root.right)  #右中左遍历
+    def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
-            root.val += self.pre
+        '''
-            self.pre = root.val
+        倒序累加替换：  
-            buildalist(root.left)
+        [2, 5, 13] -> [[2]+[1]+[0], [2]+[1], [2]] -> [20, 18, 13]
-        self.pre = 0  #记录前一个节点的数值
+        '''
-        buildalist(root)
+        self.traversal(root)
         return root
+
+    def traversal(self, root: TreeNode) -> None:
+        # 因为要遍历整棵树，所以递归函数不需要返回值
+        # Base Case
+        if not root: 
+            return None
+        # 单层递归逻辑：中序遍历的反译 - 右中左
+        self.traversal(root.right)  # 右
+
+        # 中节点：用当前root的值加上pre的值
+        root.val += self.pre.val    # 中
+        self.pre = root             
+
+        self.traversal(root.left)   # 左
 ```
 
 ## Go