diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md
index 406242a7..28425281 100644
--- a/problems/0102.二叉树的层序遍历.md
+++ b/problems/0102.二叉树的层序遍历.md
@@ -23,7 +23,7 @@
 * 111.二叉树的最小深度
 
 
-![我要打十个](https://tva1.sinaimg.cn/large/008eGmZEly1gnadnltbpjg309603w4qp.gif)
+![我要打十个](https://tva1.sinaimg.cn/large/008eGmZEly1gPnadnltbpjg309603w4qp.gif)
 
 
 
diff --git a/problems/0538.把二叉搜索树转换为累加树.md b/problems/0538.把二叉搜索树转换为累加树.md
index ec12c525..d2e98dc8 100644
--- a/problems/0538.把二叉搜索树转换为累加树.md
+++ b/problems/0538.把二叉搜索树转换为累加树.md
@@ -209,29 +209,28 @@ class Solution {
 #         self.right = right
 class Solution:
     def __init__(self):
-        self.pre = TreeNode()
+        self.count = 0
 
     def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
+        if root == None:
+            return 
         '''
         倒序累加替换：  
-        [2, 5, 13] -> [[2]+[1]+[0], [2]+[1], [2]] -> [20, 18, 13]
         '''
-        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
+        # 右
+        self.convertBST(root.right)
 
+        # 中
         # 中节点：用当前root的值加上pre的值
-        root.val += self.pre.val    # 中
-        self.pre = root             
+        self.count += root.val
 
-        self.traversal(root.left)   # 左
+        root.val = self.count 
+
+        # 左
+        self.convertBST(root.left)
+
+        return root 
+        
 ```
 
 ## Go