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+# Leetcode Insert-Into-a-Binary-Search-Tree
+
+#### 2022-07-07 08:50
+
+> ##### Algorithms:
+> #algorithm #recursion #DFS 
+> ##### Data structures:
+> #DS #binary_tree 
+> ##### Difficulty:
+> #coding_problem #difficulty-medium
+> ##### Additional tags:
+> #leetcode 
+> ##### Revisions:
+> N/A
+
+##### Related topics:
+```expander
+tag:#DFS
+```
+ 
+- [[Leetcode Binary-Tree-Inorder-Traversal]]
+- [[Leetcode Binary-Tree-Postorder-Traversal]]
+- [[Leetcode Binary-Tree-Preorder-Traversal]]
+- [[Leetcode Invert-Binary-Tree]]
+- [[Leetcode Maximum-Depth-Of-Binary-Tree]]
+- [[Leetcode Path-Sum]]
+- [[Leetcode Search-In-a-Binary-Tree]]
+- [[Leetcode Symmetric-Tree]]
+ 
+
+##### Links:
+- [Link to problem](https://leetcode.com/problems/insert-into-a-binary-search-tree/)
+___
+### Problem
+
+You are given the root node of a binary search tree (BST) and a value to insert into the tree. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST.
+
+Notice that there may exist multiple valid ways for the insertion, as long as the tree remains a BST after insertion. You can return any of them.
+
+#### Examples
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2020/10/05/insertbst.jpg)
+
+**Input:** root = [4,2,7,1,3], val = 5
+**Output:** [4,2,7,1,3,5]
+**Explanation:** Another accepted tree is:
+![](https://assets.leetcode.com/uploads/2020/10/05/bst.jpg)
+
+**Example 2:**
+
+**Input:** root = [40,20,60,10,30,50,70], val = 25
+**Output:** [40,20,60,10,30,50,70,null,null,25]
+
+**Example 3:**
+
+**Input:** root = [4,2,7,1,3,null,null,null,null,null,null], val = 5
+**Output:** [4,2,7,1,3,5]
+
+#### Constraints
+
+-   The number of nodes in the tree will be in the range `[0, 104]`.
+-   `-108 <= Node.val <= 108`
+-   All the values `Node.val` are **unique**.
+-   `-108 <= val <= 108`
+-   It's **guaranteed** that `val` does not exist in the original BST.
+
+### Thoughts
+
+> [!summary]
+> This is a #DFS problem.
+
+Seems that no one seems to care about the second way, since it's way too complex.
+
+#### DFS
+
+DFS-like solution is a simple recursion problem, using a helper function.
+
+##### Edge case:
+
+root is null, simple return a new node.
+
+##### Base cases:
+- left side is empty, and val < root->val: place it in
+- right side is empty, and val > root->val: place it in
+
+##### Pseudocode:
+- check for base cases
+- if val < root->val, insert(root->left, val)
+- vice, versa.
+
+#### BFS
+
+### Solution