Construct Binary Tree from Inorder and Postorder
Hard Problems DSA practice problem on Onlearn.
Difficulty: hard.
Topics: Constructing a Unique Binary Tree from Inorder and Postorder Traversals, Binary Tree, Tree Traversal, In-order Traversal, Postorder Traversal, Tree Construction, Recursion, Hash Map, Time Complexity, Space Complexity, inorder traversal, divide and conquer, binary tree, hash map, tree construction, postorder traversal, time complexity analysis, Construct Tree from Traversals.
Problem Statement: Given the Postorder and Inorder traversal of a Binary Tree, construct the unique Binary Tree represented by them. Input Specification: The input consists of two arrays of integers: inorder: representing the inorder traversal of the binary tree. postorder: representing the postorder traversal of the binary tree. It is guaranteed that a unique binary tree can be constructed from the given traversals. Output Specification: Return the root node of the constructed binary tree. Constraints: No explicit constraints on array sizes or value ranges are provided in the original content. Sample Test Cases: Example 1: Input: inorder = [40, 20, 50, 10, 60, 30] postorder = [40, 50, 20, 60, 30, 10] Output: (A binary tree such that its inorder traversal is [40, 20, 50, 10, 60, 30] and its postorder traversal is [40, 50, 20, 60, 30, 10]) (The tree structure is: 10 is root, 20 is left child of 10, 30 is right child of 10. 40 is left child of 20, 50 is right child of 20. 60 is left child of 30) Example 2: Input: inorder = [9, 3, 15, 20, 7] postorder = [9, 15, 7, 20, 3] Output: (A binary tree such that its inorder traversal is [9, 3, 15, 20, 7] and its postorder traversal is [9, 15, 7, 20, 3]) (The tree structure is: 3 is root, 9 is left child of 3, 20 is right child of 3. 15 is left child of 20, 7 is right child of 20)