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+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**.
+
+**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:
+    
+    
+
+**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.
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diff --git a/0784_insert-into-a-binary-search-tree/python3/solution.py b/0784_insert-into-a-binary-search-tree/python3/solution.py
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+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
+        if root == None:
+            return TreeNode(val)
+        
+        current = root
+        parent = None
+        
+        while True:
+            if val < current.val:
+                if current.left == None:
+                    current.left = TreeNode(val)
+                    break
+
+                parent = current
+                current = current.left
+            elif val >= current.val:
+                if current.right == None:
+                    current.right = TreeNode(val)
+                    break
+
+                parent = current
+                current = current.right
+        
+        return root