Tree method for conversion

Duration: 3 min

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  1. Quantitative Aptitude: Number System, Divisibility Rules, HCF LCM, Percentage, Ratio and Proportion (Ratios), Average, Profit and Loss, Simple Interest and Compound Interest, Mixture and Alligation, Partnership, Age Problems, Time and Work, Pipe and Cistern, Speed Time and Distance, Algebra, Arithmetic, Data Interpretation, Sequence and Series (Series), Logarithm, Powers and Exponents (Surds and Indices), Permutation and Combination, Probability, Speed Mathematics and Basic Mathematics, Trigonometry
  2. Logical Reasoning: Coding Decoding, Blood Relations, Directions (Direction Test), Ranking, Order and Sequence, Seating Arrangements, Puzzle Solving (Puzzles), Data Arrangement, Syllogisms, Inequality, Data Sufficiency, Deductive and Inductive Reasoning (Logical Deduction and Induction) (Prepositional Reasoning), Series (Number and Letter Series) (Numerical Relations and Reasoning), Venn Diagram, Logical Sequence Of Words, Calendar, Clock, Cubes & Dices, Non Verbal Reasoning (Spatial Aptitude) (Spatial Reasoning) (Visual Reasoning)
  3. Verbal Ability: Noun, Pronoun, Adjectives, Verb, Adverb, Tenses, Subject Verb Agreement (Verb Noun Agreement), Articles, Preposition, Conjunction, Modals, Determiner, Vocabulary, Comprehension / Reading Comprehension / Unseen Passages (Critical Reasoning) (Paragraph Questions), Cloze Test (Cloze Passage), Sentence Correction (Error Correction), Sentence Completion (Fill in the blanks), Sentence Re-arrangements (Para Jumbles) (Narrative Sequencing), Active Passive Voice (Voice), Direct and Indirect Speech, Sentence Construction
  4. Data Structures (Domain Test): Introduction to DS, Array, Stack, Queue, Linked List, Tree, Graphs, Hashing
  5. Algorithms (Domain Test): Algorithm Analysis, Sorting Algorithms, Time Complexity Analysis, Greedy Algorithms
  6. OOPs (Domain Test): Object Oriented Programming (OOPS)
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  8. SQL (Domain Test): SQL
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  11. Coding: Coding Questions by Yash Sir, C Programming by Yash Sir, Programming in C++, Java, Python for Placements by Yash Sir
  12. Written English Test: English Writing Ability
  13. Psychometric Test: Behavioral Assessment (Behavioral Competency Profiling) (Psychometric Test)
AI summary & chapters

AI Summary

An AI-generated summary of this video lecture.

This educational video features a lecture by Sanchit Jain from KnowledgeGate, focusing on converting an infix expression to postfix notation. The specific problem is from the GATE 2004 exam. The instructor systematically breaks down the expression `a + b * c - d ^ e ^ f` by constructing an expression tree based on operator precedence and associativity rules. He emphasizes that `+`, `-`, and `*` are left-associative, while `^` is right-associative, with precedence order `^` > `*` > `+` > `-`. The lecture demonstrates how to traverse the tree in post-order to derive the final postfix string.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The instructor introduces the problem statement written on the whiteboard: convert the infix expression `a + b * c - d ^ e ^ f` to postfix. He lists the operator properties, noting that `+`, `-`, `*` are left-associative and `^` is right-associative. The precedence order is established as `^` (highest), `*`, `+`, `-` (lowest). He begins constructing the expression tree by identifying the operator with the lowest precedence, which is `-`. This splits the expression into a left sub-expression `a + b * c` and a right sub-expression `d ^ e ^ f`. He then recursively analyzes the left side, identifying `+` as the root for that section, separating `a` and `b * c`. The `b * c` part is resolved with `*` as the operator. Moving to the right side, he highlights the right associativity of `^`, explaining that `d ^ e ^ f` is grouped as `d ^ (e ^ f)`. This dictates that the first `^` is the root of the right subtree, with `d` as the left child and the second `^` as the right child.

  2. 2:00 – 3:12 02:00-03:12

    The instructor completes the expression tree diagram on the board. The root is `-`. The left child is `+`, with `a` as its left child and `*` as its right child (which has `b` and `c` as children). The right child of the root is the first `^`, with `d` as its left child and the second `^` as its right child (which has `e` and `f` as children). He then performs a post-order traversal (Left, Right, Root) to generate the postfix expression. For the left subtree, the traversal yields `abc*+`. For the right subtree, the traversal yields `def^^`. Combining these parts with the root operator `-`, the final postfix expression is `abc*+def^^-`. He compares this result with the multiple-choice options, identifying Option (A) `abc x + def ^ ^ -` as the correct answer. He explicitly points out that Option (B) would be correct if `^` were left-associative, reinforcing the importance of the associativity rule.

The video provides a clear, step-by-step method for converting infix expressions to postfix notation using expression trees. The key learning points are the hierarchy of operator precedence and the impact of associativity on tree structure. Specifically, the right associativity of the exponentiation operator `^` causes the rightmost operator to become a child of the leftmost operator of the same precedence, resulting in the postfix sequence `def^^` rather than `de^f^`. This systematic approach ensures accurate conversion for complex expressions involving mixed operators.

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