diff --git a/OJ notes/pages/Leetcode 01-Matrix.md b/OJ notes/pages/Leetcode 01-Matrix.md
index f689e44..036aad9 100644
--- a/OJ notes/pages/Leetcode 01-Matrix.md	
+++ b/OJ notes/pages/Leetcode 01-Matrix.md	
@@ -1,4 +1,4 @@
-sh# Leetcode 01-Matrix
+# Leetcode 01-Matrix
 
 #### 2022-07-17 03:15
 
diff --git a/OJ notes/pages/Leetcode Merge-Two-Sorted-Lists.md b/OJ notes/pages/Leetcode Merge-Two-Sorted-Lists.md
index 290b0a4..d68343e 100644
--- a/OJ notes/pages/Leetcode Merge-Two-Sorted-Lists.md	
+++ b/OJ notes/pages/Leetcode Merge-Two-Sorted-Lists.md	
@@ -4,7 +4,7 @@
 
 ---
 ##### Algorithms:
-#algorithm #two_pointers 
+#algorithm #two_pointers #recursion
 ##### Data structures:
 #DS #linked_list 
 ##### Difficulty:
@@ -61,9 +61,51 @@ Return _the head of the merged linked list_.
 
 ### Thoughts
 
+#### Two pointers method
+
 This is a #two_pointers algorithm, I've done similar problems at leetcode's array list.
 The only thing to watch out for is when there is one list remaining, remember to add the tails.
+
+#### Recursion
+
+Very simple for recursion
+
+##### Base Case:
+- Both are nullptr -> return nullptr
+- One is nullptr -> return another
+
+##### Pseudocode:
+- Check for base case:
+- if list1->val > list2->val, list2's next should be the merged result
+- if list2->val > list1->val, list1's next should be the merged result.
+
 ### Solution
+Recursion
+```cpp
+class Solution {
+public:
+  ListNode *mergeTwoLists(ListNode *list1, ListNode *list2) {
+    // Recursion
+    // Base case, one is empty or both empty
+    if (!list1) {
+      return list2;
+    } else if (!list2) {
+      return list1;
+    }
+
+    if (list1->val > list2->val) {
+      // insert list1 after list2
+      list2->next = mergeTwoLists(list1, list2->next);
+      return list2;
+    } else {
+      list1->next = mergeTwoLists(list1->next, list2);
+      return list1;
+    }
+  }
+};
+```
+
+Two pointers
 ```cpp
 /**
  * Definition for singly-linked list.