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+Implement the `BSTIterator` class that represents an iterator over the **[in-order traversal](https://en.wikipedia.org/wiki/Tree_traversal#In-order_(LNR))** of a binary search tree (BST):
+
+*   `BSTIterator(TreeNode root)` Initializes an object of the `BSTIterator` class. The `root` of the BST is given as part of the constructor. The pointer should be initialized to a non-existent number smaller than any element in the BST.
+*   `boolean hasNext()` Returns `true` if there exists a number in the traversal to the right of the pointer, otherwise returns `false`.
+*   `int next()` Moves the pointer to the right, then returns the number at the pointer.
+
+Notice that by initializing the pointer to a non-existent smallest number, the first call to `next()` will return the smallest element in the BST.
+
+You may assume that `next()` calls will always be valid. That is, there will be at least a next number in the in-order traversal when `next()` is called.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2018/12/25/bst-tree.png)
+
+    Input
+    ["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
+    [[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]
+    Output
+    [null, 3, 7, true, 9, true, 15, true, 20, false]
+    
+    Explanation
+    BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]);
+    bSTIterator.next();    // return 3
+    bSTIterator.next();    // return 7
+    bSTIterator.hasNext(); // return True
+    bSTIterator.next();    // return 9
+    bSTIterator.hasNext(); // return True
+    bSTIterator.next();    // return 15
+    bSTIterator.hasNext(); // return True
+    bSTIterator.next();    // return 20
+    bSTIterator.hasNext(); // return False
+    
+
+**Constraints:**
+
+*   The number of nodes in the tree is in the range `[1, 105]`.
+*   `0 <= Node.val <= 106`
+*   At most `105` calls will be made to `hasNext`, and `next`.
+
+**Follow up:**
+
+*   Could you implement `next()` and `hasNext()` to run in average `O(1)` time and use `O(h)` memory, where `h` is the height of the tree?
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diff --git a/0173_binary-search-tree-iterator/python3/extra_space_precompute.py b/0173_binary-search-tree-iterator/python3/extra_space_precompute.py
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+# Time: O(N)
+# Space: O(N)
+
+# 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
+from collections import deque
+
+class BSTIterator:
+    items = None
+
+    def __init__(self, root: Optional[TreeNode]):
+        def inorder(node):
+            return inorder(node.left) + [node.val] + inorder(node.right) if node else []
+        
+        # In-order traversal of a BST gives us values in
+        # sorted order
+        self.items = deque(inorder(root))
+
+    def next(self) -> int:
+        return self.items.popleft()
+
+    def hasNext(self) -> bool:
+        return len(self.items) > 0
+
+
+# Your BSTIterator object will be instantiated and called as such:
+# obj = BSTIterator(root)
+# param_1 = obj.next()
+# param_2 = obj.hasNext()
diff --git a/0173_binary-search-tree-iterator/python3/solution.py b/0173_binary-search-tree-iterator/python3/solution.py
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