修正 0701.二叉搜索树中的插入操作，递归法版本二应该是迭代法

problems/0701.二叉搜索树中的插入操作.md

@@ -283,32 +283,10 @@ class Solution:
             return TreeNode(val)
         self.traversal(root, val)
         return root
-
 ```
 
 递归法（版本二）
 ```python
-class Solution:
-    def insertIntoBST(self, root, val):
-        if root is None:
-            return TreeNode(val)
-        parent = None
-        cur = root
-        while cur:
-            parent = cur
-            if val < cur.val:
-                cur = cur.left
-            else:
-                cur = cur.right
-        if val < parent.val:
-            parent.left = TreeNode(val)
-        else:
-            parent.right = TreeNode(val)
-        return root
-```
-
-递归法（版本三）
-```python
 class Solution:
     def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
         if root is None or root.val == val:
@@ -326,7 +304,7 @@ class Solution:
         return root
 ```
 
-递归法（版本四）
+递归法（版本三）
 ```python
 class Solution:
     def insertIntoBST(self, root, val):
@@ -340,10 +318,9 @@ class Solution:
             root.right = self.insertIntoBST(root.right, val)
 
         return root
-
 ```
 
-迭代法
+迭代法（版本一）
 ```python
 class Solution:
     def insertIntoBST(self, root, val):
@@ -366,9 +343,28 @@ class Solution:
         else:
             parent.right = node  # 将新节点连接到父节点的右子树
 
-        return root
+        return root            
+```
 
-            
+迭代法（版本二）
+```python
+class Solution:
+    def insertIntoBST(self, root, val):
+        if root is None:
+            return TreeNode(val)
+        parent = None
+        cur = root
+        while cur:
+            parent = cur
+            if val < cur.val:
+                cur = cur.left
+            else:
+                cur = cur.right
+        if val < parent.val:
+            parent.left = TreeNode(val)
+        else:
+            parent.right = TreeNode(val)
+        return root
 ```
 
 迭代法（精简）