diff --git a/problems/0235.二叉搜索树的最近公共祖先.md b/problems/0235.二叉搜索树的最近公共祖先.md
index 15ff7af4..d78db42a 100644
--- a/problems/0235.二叉搜索树的最近公共祖先.md
+++ b/problems/0235.二叉搜索树的最近公共祖先.md
@@ -265,6 +265,54 @@ class Solution:
         else: return root
 ```
 Go：
+> BSL法
+
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val   int
+ *     Left  *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+//利用BSL的性质（前序遍历有序）
+func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
+    if root==nil{return nil}
+    if root.Val>p.Val&&root.Val>q.Val{//当前节点的值大于给定的值，则说明满足条件的在左边
+        return lowestCommonAncestor(root.Left,p,q)
+    }else if root.Val<p.Val&&root.Val<q.Val{//当前节点的值小于各点的值，则说明满足条件的在右边
+        return lowestCommonAncestor(root.Right,p,q)
+    }else {return root}//当前节点的值在给定值的中间（或者等于），即为最深的祖先
+}
+```
+
+> 普通法
+
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val   int
+ *     Left  *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+//递归会将值层层返回
+func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
+    //终止条件
+    if root==nil||root.Val==p.Val||root.Val==q.Val{return root}//最后为空或者找到一个值时，就返回这个值
+    //后序遍历
+    findLeft:=lowestCommonAncestor(root.Left,p,q)
+    findRight:=lowestCommonAncestor(root.Right,p,q)
+    //处理单层逻辑
+    if findLeft!=nil&&findRight!=nil{return root}//说明在root节点的两边
+    if findLeft==nil{//左边没找到，就说明在右边找到了
+        return findRight
+    }else {return findLeft}
+}
+```
+