diff --git a/Linked List/Medium/1367.Linked List in Binary Tree b/Linked List/Medium/1367.Linked List in Binary Tree
new file mode 100644
index 0000000..9d54652
--- /dev/null
+++ b/Linked List/Medium/1367.Linked List in Binary Tree
@@ -0,0 +1,73 @@
+QUESTION :
+Given a binary tree root and a linked list with head as the first node.
+Return True if all the elements in the linked list starting from the head correspond to some downward path connected in the binary tree otherwise return False.
+In this context downward path means a path that starts at some node and goes downwards.
+
+Example 1:
+Input: head = [4,2,8], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]
+Output: true
+Explanation: Nodes in blue form a subpath in the binary Tree.
+
+Example 2:
+Input: head = [1,4,2,6], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]
+Output: true
+
+Example 3:
+Input: head = [1,4,2,6,8], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]
+Output: false
+Explanation: There is no path in the binary tree that contains all the elements of the linked list from head.
+
+Constraints:
+1 <= node.val <= 100 for each node in the linked list and binary tree.
+The given linked list will contain between 1 and 100 nodes.
+The given binary tree will contain between 1 and 2500 nodes.
+
+C++ SOLUTION :
+
+class Solution {
+public:
+ bool isSubPath(ListNode* head, TreeNode* root) {
+ if(root){
+ if(isEqual(head, root)){
+ return true;
+ }
+ return isSubPath(head, root->left) || isSubPath(head, root->right);
+ }
+ return false;
+ }
+ bool isEqual(ListNode* head, TreeNode* root){
+ if(!head){
+ return true;
+ }
+ if(!root){
+ return false;
+ }
+ return head->val == root->val && (isEqual(head->next, root->left) || isEqual(head->next, root->right));
+ }
+};
+
+PYTHON SOLUTION :
+
+def isSubPath(self, head: ListNode, root: TreeNode) -> bool:
+ target = ""
+ while head:
+ target = target + str(head.val)
+ head = head.next
+
+ def dfs(root, path):
+ if target in path:
+ return True
+
+ if root.left:
+ ans = dfs(root.left, path + str(root.left.val))
+ if ans == True:
+ return True
+
+ if root.right:
+ ans = dfs(root.right, path + str(root.right.val))
+ if ans == True:
+ return True
+
+ return False
+
+ return dfs(root, str(root.val))
diff --git a/README.md b/README.md
index 79c670f..3132639 100644
--- a/README.md
+++ b/README.md
@@ -1,9 +1,16 @@
# LeetCode
-* 贵有恒，何必三更起五更睡；最无益，只怕一日暴十寒。
-* [我的博客](http://blog.csdn.net/c406495762 "悬停显示")
-* 我的个人网站：[http://cuijiahua.com/](http://cuijiahua.com/ "悬停显示")
-* 分享技术，乐享生活：Jack Cui公众号每周五推送“程序员欢乐送”系列资讯类文章，欢迎您的关注！
-
+
+原创文章每周最少两篇，**后续最新文章**会在[【公众号】](https://cuijiahua.com/wp-content/uploads/2020/05/gzh-w.jpg)首发，视频[【B站】](https://space.bilibili.com/331507846)首发，大家可以加我[【微信】](https://cuijiahua.com/wp-content/uploads/2020/05/gzh-w.jpg)进**交流群**，技术交流或提意见都可以，欢迎**Star**！
+
+
+ 
+ 
+ 
+ 
+ 
+ 
+ 
+
## 帮助文档
@@ -98,7 +105,7 @@
| 290 | Easy | Word Pattern.mde | [Python](https://github.com/Jack-Cherish/LeetCode/blob/master/Hash%20Table/Easy/290.Word%20Pattern.md "悬停显示") | no | [思路讲解](https://github.com/Jack-Cherish/LeetCode/blob/master/Hash%20Table/Easy/290.Word%20Pattern.md "悬停显示") |
# 剑指Offer（已完成）
-* 贵有恒，何必三更起五更睡；最无益，只怕一日暴十寒。
+
* [个人网站](http://cuijiahua.com/ "悬停显示")
### 链表(8道)：
@@ -192,3 +199,9 @@
* [剑指Offer（五十四）：字符流中第一个不重复的字符](http://cuijiahua.com/blog/2018/01/basis_54.html "悬停显示")
* [剑指Offer（六十三）：数据流中的中位数](http://cuijiahua.com/blog/2018/02/basis_63.html "悬停显示")
* [剑指Offer（六十四）：滑动窗口的最大值](http://cuijiahua.com/blog/2018/02/basis_64.html "悬停显示")
+
+更多精彩，敬请期待！
+
+
+
+
+