From e06b12d5919c2e4f2c19a90ecd360f3cefdc03b1 Mon Sep 17 00:00:00 2001
From: Manasvi Reddy <manasvireddyk25@gmail.com>
Date: Sun, 5 Jul 2026 22:50:20 -0400
Subject: [PATCH] Done Trees-1

---
 Problem1.py | 39 +++++++++++++++++++++++++++++++++++++++
 Problem2.py | 53 +++++++++++++++++++++++++++++++++++++++++++++++++++++
 2 files changed, 92 insertions(+)
 create mode 100644 Problem1.py
 create mode 100644 Problem2.py

diff --git a/Problem1.py b/Problem1.py
new file mode 100644
index 00000000..bf7f9b16
--- /dev/null
+++ b/Problem1.py
@@ -0,0 +1,39 @@
+#https://leetcode.com/problems/validate-binary-search-tree/
+# Time Complexity: O(n), we visit every node exactly once
+# Space Complexity: O(h), h is the height of the tree, due to recursion stack
+# Approach: do an in-order traversal (left, current, right). 
+#In a valid BST, this visits values in strictly increasing order, so we just compare each node to the one visited right before it.
+
+# 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:
+       
+
+        self.flag = True      # assume valid until we find proof otherwise
+        self.prev = None      # tracks the previously visited node in the in-order walk
+
+        def helper(currentNode):
+            # base case: we've gone past a leaf, nothing here to check
+            if currentNode is None:
+                return
+
+            # step 1: fully process the entire left side first
+            helper(currentNode.left)
+
+            # step 2: now check the current node against the last one we visited
+            if self.prev is not None and self.prev.val >= currentNode.val:
+                self.flag = False
+
+            # step 3: this node is now "the previous one" for whatever comes next
+            self.prev = currentNode
+
+            # step 4: now process the entire right side
+            helper(currentNode.right)
+
+        helper(root)        # kick off the traversal starting at the actual tree root
+        return self.flag    # report whether any violation was found
\ No newline at end of file
diff --git a/Problem2.py b/Problem2.py
new file mode 100644
index 00000000..21156aa9
--- /dev/null
+++ b/Problem2.py
@@ -0,0 +1,53 @@
+#https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
+# Time Complexity: O(n), We visit every node exactly once and each visit does constant work using dictionary lookup and a moving pointer
+# Space Complexity: O(n), We use a dictionary of size n to store index positions and the recursion stack can go up to depth n in the worst case
+# Approach: The first value in preorder is always the root of the current subtree so we read it using a pointer that moves forward
+# We use a dictionary to instantly find where that root sits in inorder so we know how many nodes go to the left and right
+# We always build the left subtree before the right subtree because reading preorder forward gives root then left side then right side
+
+# 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]:
+        # this dictionary will store the index of every value in inorder so we never have to search for it later
+        self.map = {}
+
+        # this pointer starts at the first index of preorder and will move forward as we build each root
+        self.preorder_index = 0
+
+        # fill the dictionary once so lookups later take constant time
+        for i in range(len(inorder)):
+            self.map[inorder[i]] = i
+
+        # start building the tree using the full range of inorder, from index 0 to the last index
+        return self.helper(preorder, 0, len(inorder) - 1)
+
+    def helper(self, preorder, start, end):
+        # if start has gone past end it means there are no elements left in this range so there is nothing to build
+        if start > end:
+            return None
+
+        # the value at our current pointer position in preorder is always the root of this subtree
+        root_value = preorder[self.preorder_index]
+
+        # move the pointer one step forward so the next call reads the correct next value
+        self.preorder_index += 1
+
+        # look up where this root value sits in inorder so we know how to split left and right parts
+        root_index = self.map[root_value]
+
+        # create the node for this root value
+        node = TreeNode(root_value)
+
+        # build the left subtree first since reading preorder forward gives left side nodes before right side nodes
+        node.left = self.helper(preorder, start, root_index - 1)
+        # only after the left subtree is fully built do we build the right subtree
+        node.right = self.helper(preorder, root_index + 1, end)
+
+        # return this node with both children attached back to whoever called this
+        return node
+        
\ No newline at end of file