Add implementation for leetcode problems 98, 105

leetcode_105.py

@@ -0,0 +1,108 @@
+#!/usr/bin/env python3
+# -*- coding: utf-8 -*-
+"""
+Created on Sun Mar 15 00:00:00 2026
+
+@author: rishigoswamy
+
+    LeetCode 105: Construct Binary Tree from Preorder and Inorder Traversal
+    Link: https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
+
+    Problem:
+    Given two integer arrays preorder and inorder where preorder is the preorder
+    traversal of a binary tree and inorder is the inorder traversal of the same tree,
+    construct and return the binary tree.
+
+    Note: You may assume that duplicates do not exist in the tree.
+
+    Approach:
+    Use a hashmap for O(1) inorder index lookup and a global index pointer into preorder.
+    The first element of preorder is always the root. Its position in inorder splits
+    the tree into left and right subtrees.
+
+    1️⃣ Build inorderMap: value → index for O(1) lookup.
+    2️⃣ self.idx tracks the current root in preorder (starts at 0).
+    3️⃣ createTree(left, right) defines the valid inorder window.
+    4️⃣ Pick preorder[self.idx] as root, increment self.idx.
+    5️⃣ Recurse left on inorder[left : nodeIdx-1], then right on inorder[nodeIdx+1 : right].
+
+    // Time Complexity : O(n)
+        Each node is visited once; hashmap gives O(1) index lookup.
+    // Space Complexity : O(n)
+        O(n) for the hashmap + O(h) recursive call stack (h = height of tree).
+
+"""
+
+from typing import List, Optional
+
+# 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]:
+        self.inorderMap = {val: idx for idx, val in enumerate(inorder)}  # Map value to index for O(1) lookup
+        self.idx = 0
+        return self.createTree(preorder, 0, len(inorder) - 1)
+
+    def createTree(self, preOrder, left, right):
+        if left > right:
+            return None
+
+        nodeValue = preOrder[self.idx]
+        self.idx += 1
+
+        nodeIdx = self.inorderMap[nodeValue]
+
+        node = TreeNode(nodeValue)
+
+        node.left = self.createTree(preOrder, left, nodeIdx - 1)
+        node.right = self.createTree(preOrder, nodeIdx + 1, right)
+
+        return node
+
+    '''
+    def createTree(self, preorder: List[int], inorder: List[int], inorderMap: dict) -> Optional[TreeNode]:
+        if not preorder or not inorder:
+            return None  # Base case: stop recursion if preorder or inorder list is empty
+
+        root_val = preorder[0]
+        root = TreeNode(root_val)  # Create the root node with the first value in preorder
+
+        indexroot = -1
+        for item in inorder:
+            indexroot += 1
+            if item == root_val:
+                break
+
+        leftin = inorder[:indexroot]
+        rightin = inorder[indexroot + 1:]
+
+        leftpre = preorder[1: len(leftin) + 1]
+        rightpre = preorder[len(leftin) + 1:]
+
+        root.left = self.createTree(leftpre, leftin, inorderMap) if leftin else None
+        root.right = self.createTree(rightpre, rightin, inorderMap) if rightin else None
+
+        return root
+
+    def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
+        inorderMap = {val: idx for idx, val in enumerate(inorder)}  # Map value to index for O(1) lookup
+        return self.createTree(preorder, inorder, inorderMap)
+    '''
+
+    '''
+    def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
+        if not preorder:
+            return None
+
+        root = TreeNode(preorder[0])
+        mid = inorder.index(preorder[0])
+        root.left = self.buildTree(preorder[1:mid+1], inorder[:mid])
+        root.right = self.buildTree(preorder[mid+1:], inorder[mid+1:])
+        return root
+    '''

leetcode_98.py

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+#!/usr/bin/env python3
+# -*- coding: utf-8 -*-
+"""
+Created on Sat Mar 15 00:00:00 2026
+
+@author: rishigoswamy
+
+    LeetCode 98: Validate Binary Search Tree
+    Link: https://leetcode.com/problems/validate-binary-search-tree/
+
+    Problem:
+    Given a binary tree, determine if it is a valid binary search tree (BST).
+    A BST is defined as follows:
+    - The left subtree of a node contains only nodes with keys less than the node's key.
+    - The right subtree of a node contains only nodes with keys greater than the node's key.
+    - Both the left and right subtrees must also be binary search trees.
+
+    Approach:
+    Inorder traversal with a running prev pointer.
+    In a valid BST, inorder traversal always yields a strictly increasing sequence.
+    Track self.prev (initialized to -inf) and self.flag for early termination.
+
+    1️⃣ Recurse left subtree.
+    2️⃣ Check if self.prev >= current node's value → invalid, set flag = False.
+    3️⃣ Update self.prev to current node's value.
+    4️⃣ Recurse right subtree.
+
+    // Time Complexity : O(n)
+        Every node is visited once.
+    // Space Complexity : O(h)
+        Recursive call stack depth equals the height of the tree.
+
+"""
+
+import math
+from typing import Optional
+
+# 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
+        self.prev = -math.inf
+        def validate(root):
+            if not self.flag: # This is optimization. Can be removed
+                return False
+            if not root:
+                return True
+            validate(root.left)
+            if self.prev >= root.val:
+                self.flag = False
+            self.prev = root.val
+            validate(root.right)
+
+        validate(root)
+        return self.flag
+
+
+        '''
+        self.prev = -math.inf
+        def validate(root):
+
+            if not root:
+                return True
+            left = True
+            left = validate(root.left)
+
+            if not left:
+                return False
+            if self.prev >= root.val:
+                self.flag = False
+                return False
+            self.prev = root.val
+            right = validate(root.right)
+
+            return left and right
+
+        return validate(root)
+
+        '''
+        '''
+        if not root:
+            return True
+
+        def validate(root, low = -math.inf, high = math.inf):
+            if not root:
+                return True
+            if root.val <= low or root.val >= high:
+                return False
+
+            return validate(root.left, low, root.val) and validate(root.right,
+                root.val, high)
+        return validate(root)
+        '''
+        '''
+        if not root:
+            return True
+
+
+        def traverse(root, res):
+            if root.left:
+                traverse(root.left, res)
+            res.append(root.val)
+            if root.right:
+                traverse(root.right, res)
+
+        res = []
+        traverse(root, res)
+
+        for i in range(1, len(res)):
+            if res[i-1] >= res[i]:
+                return False
+
+        return True
+        '''