diff --git a/Problem1.py b/Problem1.py
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+# https://leetcode.com/problems/validate-binary-search-tree/
+# 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 isValidBST(self, root: Optional[TreeNode]) -> bool:
+        # prev = None
+        # isValid = True
+        # def helper(root):
+        #     nonlocal prev, isValid
+        #     if root == None:
+        #         return
+            
+        #     helper(root.left)
+        #     if prev and prev.val >= root.val:
+        #         isValid = False
+        #     prev = root
+        #     helper(root.right)
+        
+        # helper(root)
+        # return isValid
+
+
+
+        """
+        Time Complexity: O(N)
+        - N is the total number of nodes in the binary tree.
+        - We visit every single node exactly once to check its value against the boundaries.
+        
+        Space Complexity: O(N) in the worst case, O(log N) in the best/average case
+        - The space is consumed by the recursion call stack.
+        - Worst case: O(N) if the tree is completely unbalanced (like a linked list).
+        - Best/Average case: O(log N) if the tree is balanced, where the height of 
+          the tree dictates the maximum depth of the recursive stack.
+        """
+        
+        # Helper function that passes the valid boundary limits down the tree
+        def helper(node, minVal, maxVal):
+            # Base Case: Reaching the end of a branch (an empty node) means 
+            # this path is valid so far.
+            if node == None:
+                return True
+            
+            # Logic: The current node must be strictly greater than the minimum boundary.
+            # (minVal is updated when we move to the right child of an ancestor).
+            if minVal is not None and node.val <= minVal:
+                return False
+            
+            # Logic: The current node must be strictly less than the maximum boundary.
+            # (maxVal is updated when we move to the left child of an ancestor).
+            if maxVal is not None and node.val >= maxVal:
+                return False
+            
+            # Recursive Step:
+            # 1. Check the left subtree: The current node's value becomes the new strict upper limit (maxVal).
+            # 2. Check the right subtree: The current node's value becomes the new strict lower limit (minVal).
+            # Both sides must return True for the entire tree to be valid.
+            return helper(node.left, minVal, node.val) and helper(node.right, node.val, maxVal)
+            
+        # Start at the root. Initially, there are no upper or lower boundaries.
+        return helper(root, None, None)
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diff --git a/Problem2.py b/Problem2.py
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+# https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
+# 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 buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
+        """
+        Logic:
+        1. The first element in the `preorder` array is always the root of the current tree/subtree.
+        2. We locate this root value in the `inorder` array. 
+        3. In an inorder traversal, everything to the left of the root belongs to the left subtree, 
+           and everything to the right belongs to the right subtree.
+        4. We use the size of the left subtree (from `inorder`) to figure out which elements in 
+           `preorder` belong to the left vs. right subtrees.
+        5. We recursively repeat this process for the left and right subtrees.
+
+        Time Complexity: O(N^2) in the worst case (where N is the number of nodes). 
+                         For each node, we are searching for its index in `inorder` which takes O(N) time.
+                         Additionally, list slicing takes O(N) time. In a skewed tree, the recursion depth 
+                         is N, resulting in N * O(N) = O(N^2) time.
+        
+        Space Complexity: O(N^2) in the worst case. At each recursive call, we are creating new arrays 
+                          via list slicing (`inLeft`, `inRight`, etc.). For a skewed tree with depth N, 
+                          creating these slices repeatedly consumes O(N^2) memory. The recursion stack 
+                          also takes O(N) space.
+        """
+        
+        # Base case: If there are no elements left, the subtree is empty
+        if len(preorder) == 0:
+            return None
+
+        # The root of the current subtree is always the first element in preorder
+        rootVal = preorder[0]
+        
+        # Find the index of this root in the inorder list
+        inorderRootIdx = -1
+        for i in range(len(inorder)):
+            if inorder[i] == rootVal:
+                inorderRootIdx = i
+                break
+
+        # Split the inorder list into left and right subtrees
+        inLeft = inorder[0:inorderRootIdx]
+        inRight = inorder[(inorderRootIdx + 1):len(inorder)]
+        
+        # Split the preorder list into left and right subtrees
+        # The number of elements in the left subtree is len(inLeft). 
+        # We take that many elements from preorder (skipping the first element, which is the root).
+        preLeft = preorder[1:(1 + len(inLeft))]
+        preRight = preorder[(1 + len(inLeft)):len(preorder)]
+
+        # Construct the root node
+        root = TreeNode(rootVal)
+        
+        # Recursively construct the left and right subtrees using the sliced arrays
+        root.left = self.buildTree(preLeft, inLeft)
+        root.right = self.buildTree(preRight, inRight)
+
+        return root
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