diff --git a/problems/0701.二叉搜索树中的插入操作.md b/problems/0701.二叉搜索树中的插入操作.md
index 5e9fbdfe..df6a3954 100644
--- a/problems/0701.二叉搜索树中的插入操作.md
+++ b/problems/0701.二叉搜索树中的插入操作.md
@@ -280,7 +280,7 @@ class Solution:
 
         # 返回更新后的以当前root为根节点的新树
         return roo
-``` 
+```
 
 **递归法** - 无返回值
 ```python
@@ -308,7 +308,7 @@ class Solution:
             return
         __traverse(root, val)
         return root
-``` 
+```
 
 **递归法** - 无返回值 - another easier way
 ```python
@@ -378,7 +378,7 @@ func insertIntoBST(root *TreeNode, val int) *TreeNode {
     }
     return root
 }
-``` 
+```
 
 迭代法 
 
@@ -520,5 +520,71 @@ var insertIntoBST = function (root, val) {
 };
 ```
 
+## TypeScript
+
+> 递归-有返回值
+
+```typescript
+function insertIntoBST(root: TreeNode | null, val: number): TreeNode | null {
+    if (root === null) return new TreeNode(val);
+    if (root.val > val) {
+        root.left = insertIntoBST(root.left, val);
+    } else {
+        root.right = insertIntoBST(root.right, val);
+    }
+    return root;
+};
+```
+
+> 递归-无返回值
+
+```typescript
+function insertIntoBST(root: TreeNode | null, val: number): TreeNode | null {
+    if (root === null) return new TreeNode(val);
+    function recur(root: TreeNode | null, val: number) {
+        if (root === null) {
+            if (parentNode.val > val) {
+                parentNode.left = new TreeNode(val);
+            } else {
+                parentNode.right = new TreeNode(val);
+            }
+            return;
+        }
+        parentNode = root;
+        if (root.val > val) recur(root.left, val);
+        if (root.val < val) recur(root.right, val);
+    }
+    let parentNode: TreeNode = root;
+    recur(root, val);
+    return root;
+};
+```
+
+> 迭代法
+
+```typescript
+function insertIntoBST(root: TreeNode | null, val: number): TreeNode | null {
+    if (root === null) return new TreeNode(val);
+    let curNode: TreeNode | null = root;
+    let parentNode: TreeNode = root;
+    while (curNode !== null) {
+        parentNode = curNode;
+        if (curNode.val > val) {
+            curNode = curNode.left
+        } else {
+            curNode = curNode.right;
+        }
+    }
+    if (parentNode.val > val) {
+        parentNode.left = new TreeNode(val);
+    } else {
+        parentNode.right = new TreeNode(val);
+    }
+    return root;
+};
+```
+
+
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>