From d60fd1c180885c767a32134d76662ab51ca7fcc0 Mon Sep 17 00:00:00 2001
From: Shinjanee Gupta <shinjaneegupta@gmail.com>
Date: Mon, 23 Feb 2026 21:59:04 -0800
Subject: [PATCH] Completed Trees-1

---
 constructBinaryTree.py | 38 ++++++++++++++++++++++++++++++++++++++
 validatebst.py         | 25 +++++++++++++++++++++++++
 2 files changed, 63 insertions(+)
 create mode 100644 constructBinaryTree.py
 create mode 100644 validatebst.py

diff --git a/constructBinaryTree.py b/constructBinaryTree.py
new file mode 100644
index 00000000..d09eb53a
--- /dev/null
+++ b/constructBinaryTree.py
@@ -0,0 +1,38 @@
+# Time Complexity : O(n)
+# Space Complexity : O(n)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach : elements to the left and right of root in inorder array are used to form the left and right subtree.
+
+# 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 __init__(self):
+        self.preorder_idx = 0
+
+    def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
+        def tree(left, right):
+            if left > right:
+                return None
+
+            root_val = preorder[self.preorder_idx]
+            root = TreeNode(root_val)
+            self.preorder_idx += 1
+
+            root.left = tree(left, inorder_idx_map[root_val] - 1)
+            root.right = tree(inorder_idx_map[root_val] + 1, right)
+
+            return root
+
+
+        inorder_idx_map = {}
+
+        for idx, val in enumerate(inorder):
+            inorder_idx_map[val] = idx
+
+        return tree(0, len(preorder) - 1)
+        
\ No newline at end of file
diff --git a/validatebst.py b/validatebst.py
new file mode 100644
index 00000000..4690ce39
--- /dev/null
+++ b/validatebst.py
@@ -0,0 +1,25 @@
+# Time Complexity : O(n)
+# Space Complexity : O(n)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach: Recursively ensure each node’s value lies within its allowed low and high range.
+
+# 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 validate(self, root: Optional[TreeNode], low: int, high: int) -> bool:
+        if not root:
+            return True
+        
+        if root.val <= low or root.val >= high:
+            return False
+        
+        return self.validate(root.left, low, root.val) and self.validate(root.right, root.val, high)
+
+    def isValidBST(self, root: Optional[TreeNode]) -> bool:
+        return self.validate(root, -float('inf'), float('inf'))
+        
\ No newline at end of file