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 a4164b2c..fdfde822 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
@@ -835,35 +835,35 @@ public:
 python代码：
 
 ```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 rightSideView(self, root: TreeNode) -> List[int]:
         if not root:
             return []
-
-        # deque来自collections模块，不在力扣平台时，需要手动写入
-        # 'from collections import deque' 导入
-        # deque相比list的好处是，list的pop(0)是O(n)复杂度，deque的popleft()是O(1)复杂度
-
-        quene = deque([root])
-        out_list = []
-
-        while quene:
-            # 每次都取最后一个node就可以了
-            node = quene[-1]
-            out_list.append(node.val)
-
-            # 执行这个遍历的目的是获取下一层所有的node
-            for _ in range(len(quene)):
-                node = quene.popleft()
+        
+        queue = collections.deque([root])
+        right_view = []
+        
+        while queue:
+            level_size = len(queue)
+            
+            for i in range(level_size):
+                node = queue.popleft()
+                
+                if i == level_size - 1:
+                    right_view.append(node.val)
+                
                 if node.left:
-                    quene.append(node.left)
+                    queue.append(node.left)
                 if node.right:
-                    quene.append(node.right)
-
-        return out_list
-
-# 执行用时：36 ms, 在所有 Python3 提交中击败了89.47%的用户
-# 内存消耗：14.6 MB, 在所有 Python3 提交中击败了96.65%的用户
+                    queue.append(node.right)
+        
+        return right_view
 ```