BST Practice Question
Duration: 7 min
This video lesson is available to enrolled students.
Inside: a video lesson and guided study material.
Module outline
- About the course: About the Course
- Mathematical Aptitude: Number System, HCF LCM, Divisibility Rules, Percentage, Ratio and Proportion (Ratios), Average, Profit and Loss, Time and Work, Mensuration and Geometry, Simple Interest and Compound Interest, Speed Time and Distance, Age Problems, Mixture and Alligation, Algebra, Powers and Exponents (Surds and Indices), Pipe and Cistern, Trigonometry, Statistics, Data Interpretation, Arithmetic
- Logical Reasoning: Series (Number and Letter Series) (Numerical Relations and Reasoning), Coding Decoding, Analogy, Blood Relations, Directions (Direction Test), Ranking, Order and Sequence, Non Verbal Reasoning (Spatial Aptitude) (Spatial Reasoning) (Visual Reasoning), Seating Arrangements, Syllogisms, Venn Diagram, Puzzle Solving (Puzzles), Logical Sequence Of Words, Deductive and Inductive Reasoning (Logical Deduction and Induction) (Prepositional Reasoning), Analytical Reasoning (Counting Figures Reasoning), Cubes & Dices, Calendar, Clock
- General Knowledge: Physics, Chemistry, Biology, History, Geography, Polity, Indian Economy, GK- Current Affairs, Static General Knowledge
- Art of Teaching: Nature, Objectives & Characteristics of Teaching, Learners & Learning Process, Factors Affecting Teaching, Methods of Teaching, Classroom Management & Communication, Teaching-Learning Aids & ICT Integration, Evaluation, Assessment & Measurement, Professional Development & Ethics, Policies, Acts & Contemporary Issues, Curriculum & School Environment, Guidance, Counselling & Mental Health, Inclusive Education & Special Needs, Micro-Teaching & Teaching Skills, Lesson Planning
- Environmental Science: Environment
- Unit 1 | Digital Electronics: Digital Systems & Boolean Basics, Logic Gates & Hardware, Boolean Expression, Boolean Minimization, Combinational Circuit, Sequential Circuits, Number System, Number Representation
- Unit 2 | COA: Basics of COA, Floating Point Rep, Cache Memory Organization, Input Output Organisation, Instr Formats & Modes, Control Unit Design, Register Transfer and Microoperations, Programming the Basic Computer
- Unit 3 | C Programming: C Fundamentals, Control Flow, Functions, Arrays & Pointers, Storage Classes, Structures & Enums, DMA, Macros, Scoping & File Handling
- Unit 3 | Data Structures: Introduction to DS, Array, Stack, Queue, Linked List, Tree, Graphs, Hashing
- Unit 4 | Algorithms: Algorithm Analysis, Time Complexity Analysis, Sorting Algorithms, Dynamic Programming, Minimum Spanning Trees, Shortest Path Algos, Advanced Algorithms
- Unit 5 | Operating System: Introduction to OS, Process Management, CPU Scheduling, Process Synchronization, Threads & Process Creation, Deadlock, Memory Management, Virtual Memory, Disc Scheduling, File Management
- Unit 6 | Database Management System: Basics of DBMS, ER Diagram, Relational Model & Functional Dependencies, Keys & Integrity Constraints, Normalization (1NF - BCNF), Decomposition Properties & 4NF, File Organization & Indexing, Relational Algebra, SQL, Transaction Management, Concurrency Control
- Unit 7 | Computer Networks: Introduction to CN, Data Communication, DLL: Access Control, DLL: Flow Control, DLL: Error Control, DLL: Framing, Data Link Layer - Ethernet, Net Layer: IPv4 & Proto, Net Layer: IP Addressing, Net Layer:Routing Protocol, Transport Layer Services, TL: Congestion & UDP, Application Layer
- Unit 8 | Software Engineering: Fundamentals of Software Engineering, Software Requirements and Quality Assurance, Software Design, Estimation and Metrics, Software Testing, Software Maintenance and Configuration Management
- Unit 9 | Object Oriented Programming: OOPS with C++
- Unit 10 | Web Development: HTML Basics, CSS, JavaScript-Basics, XML
- Unit 11 | Theory Of Computation: Introduction to TOC, Deterministic FA (DFA), Non-Deterministic FA, Regular Expressions, Grammar, Regular Language Properties, Moore & Mealy Machines, Pushdown Automata & CFG, Turing Machines
- Unit 12 | IoT: Emerging Tech Trends, Cloud Computing
- Unit 13 | Artificial Intelligence: Approaches to AI, Search Algorithms, Game Playing, Knowledge Representation, Natural Language Processing, Artificial Neural Networks, Genetic Algorithms
- Unit 14 | E-Commerce: Electronic Data Interchange and E-Commerce, Role of Internet and Web Services, Security in Mobile and Wireless Computing, Physical Security and Cryptography Basics, Cyber Safety and Security
- Unit 15 | Multimedia: Multimedia
- Unit 16 | Python: Introduction to Python, Tokens, Variables, Data Types, I/O, Operators & Expressions, Control Flow & Looping, Collections & Strings, Functions & Modules, File Handling & Data Analysis, DSA using Python
- Full Length Mock Tests — General Paper:
- Full Length Mock Tests — Computer Science (Paper II):
AI summary & chapters
AI Summary
An AI-generated summary of this video lecture.
This lecture segment addresses a binary search tree (BST) practice question: how many distinct BSTs can be constructed with 3 distinct keys. The instructor presents the multiple-choice options (4, 5, 6, or 9), then works through a concrete example using keys 1, 2, and 3. By systematically enumerating all possible insertion sequences (permutations) and drawing the resulting tree structures, the instructor demonstrates that each permutation yields a unique BST shape. The six distinct trees are numbered and displayed, leading to the answer c) 6. In the final segment, the instructor begins introducing a more general combinatorial approach using a root node K with left and right subtrees, writing partial calculations such as 1*1=1 and 2*1=2, suggesting a recursive counting method for larger N.
Chapters
0:00 – 2:00 00:00-02:00
The slide displays the question 'Q how many distinct BST can be constructed with 3 distinct keys?' in orange text, with answer options a) 4, b) 5, c) 6, d) 9. Red hand-drawn underlines emphasize key terms: 'distinct', 'BST can be constructed', and '3 distinct keys'. The instructor then writes 'Example -> N = 3' and the key set '1,2,3', beginning to draw the first BST structure by placing node 1 as root, then adding node 2 as a right child, and node 3 as a right child of 2, forming a right-leaning chain.
2:00 – 5:00 02:00-05:00
The instructor continues drawing additional BST structures for different insertion sequences. The first tree (1,2,3) is a right-skewed chain tagged with circled '1'. The second tree (1,3,2) shows root 1 with right child 3 and left child 2 beneath it, tagged '2'. A balanced tree for sequence 2,1,3 is drawn with root 2 and children 1 (left) and 3 (right). A right-skewed tree for 3,1,2 and a left-skewed tree for 3,2,1 are also drawn. Numbered labels from 1 to 6 appear below each tree, confirming six distinct configurations and supporting answer option c) 6.
5:00 – 7:29 05:00-07:29
The board now shows all six hand-drawn BSTs grouped by root: sequences 1,2,3 and 1,3,2 under root 1; 2,1,3 and 2,3,1 under root 2 with children 1 and 3; and 3,1,2 and 3,2,1 under root 3. A bottom-left diagram introduces a general node 'K' branching to two circles, one labeled 'No of possible combination structures.' The instructor writes partial calculations including a crossed-out '1 * 2', then '1 * 1 = 1' and '2 * 1 = 2', hinting at a recursive formula for counting BSTs by multiplying left and right subtree possibilities. The segment ends before the full general solution is completed.
The core teaching objective is to connect permutation-based enumeration with BST structure uniqueness. The instructor uses N=3 as a tractable example: all 6 permutations of keys {1,2,3} produce distinct tree shapes because BST insertion order determines structure. This concrete demonstration validates the answer c) 6 and builds intuition for why the count equals the number of permutations for small N. The transition to a general method via node K with left/right subtree multiplication suggests the Catalan number recurrence C(n) = sum_{i=0}^{n-1} C(i)*C(n-1-i), though this is only partially developed in the sampled frames. Students should note that for N=3, the six trees correspond exactly to 3! = 6 permutations, but this equivalence does not hold for larger N where different insertion sequences can yield the same shape.