GATE CS 2020 Set 1 — Question 15

MCQ+1 / -0.33MediumTree TraversalsTreesProgramming & Data StructuresBinary Search Trees

Programming & Data Structures → Trees → Binary Search Trees

Last updated 5 September 2026

Question

The preorder traversal of a binary search tree is 15, 10, 12, 11, 20, 18, 16, 19. Which one of the following is the postorder traversal of the tree?

Correct answer(B) 11, 12, 10, 16, 19, 18, 20, 15SolutionGiven Preorder: 15, 10, 12, 11, 20, 18, 16, 19.In a BST, the first element of preorder is the root. Elements smaller than the root form the left subtree, and elements larger form the right subtree.1.Root: 152.Left Subtree (elements < 15): {10, 12, 11} Root: 10 Right Subtree (> 10): {12, 11} Root: 12 Left Child (< 12): 11 Structure: 10 -> Right(12 -> Left(11))3. Right Subtree (elements > 15): {20, 18, 16, 19} Root: 20 Left Subtree (< 20): {18, 16, 19} Root: 18 Left Child (< 18): 16 Right Child (> 18): 19 Structure: 20 -> Left(18 -> Left(16), Right(19))Tree Structure: 15 / \ 10 20 \ / 12 18 / / \ 11 16 19Postorder Traversal (Left, Right, Root): Left Subtree of 15: (Left of 10 is null), (Right of 10 is Postorder(12, 11) -> 11, 12), Root 10. -> 11, 12, 10 Right Subtree of 15: (Left of 20 is Postorder(18, 16, 19) -> 16, 19, 18), (Right of 20 is null), Root 20. -> 16, 19, 18, 20 Root 15.Combined: 11, 12, 10, 16, 19, 18, 20, 15.Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.

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Correct answer(B) 11, 12, 10, 16, 19, 18, 20, 15SolutionGiven Preorder: 15, 10, 12, 11, 20, 18, 16, 19.In a BST, the first element of preorder is the root. Elements smaller than the root form the left subtree, and elements larger form the right subtree.1.Root: 152.Left Subtree (elements < 15): {10, 12, 11} Root: 10 Right Subtree (> 10): {12, 11} Root: 12 Left Child (< 12): 11 Structure: 10 -> Right(12 -> Left(11))3. Right Subtree (elements > 15): {20, 18, 16, 19} Root: 20 Left Subtree (< 20): {18, 16, 19} Root: 18 Left Child (< 18): 16 Right Child (> 18): 19 Structure: 20 -> Left(18 -> Left(16), Right(19))Tree Structure: 15 / \ 10 20 \ / 12 18 / / \ 11 16 19Postorder Traversal (Left, Right, Root): Left Subtree of 15: (Left of 10 is null), (Right of 10 is Postorder(12, 11) -> 11, 12), Root 10. -> 11, 12, 10 Right Subtree of 15: (Left of 20 is Postorder(18, 16, 19) -> 16, 19, 18), (Right of 20 is null), Root 20. -> 16, 19, 18, 20 Root 15.Combined: 11, 12, 10, 16, 19, 18, 20, 15.

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