BST Practice Question

Duration: 2 min

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Inside: a video lesson and guided study material.

Module outline

  1. C: C Programming by Yash Sir
  2. C++: Programming in C++
  3. Java: Java
  4. Python: Python for Placements by Yash Sir
  5. OOPs Concept: Object Oriented Programming (OOPS)
  6. Pseudo Code: Pseudo Codes
  7. Data Structure: Introduction to DS, Array, Stack, Queue, Linked List, Tree, Graphs, Hashing
  8. Algorithm: Algorithm Analysis, Sorting Algorithms, Greedy Algorithms, Dynamic Programming, Minimum Spanning Trees, Shortest Path Algos, Time Complexity Analysis, Advanced Algorithms
  9. DSA Implementation: DSA using Python
  10. Competitive Coding: Coding Questions by Yash Sir, Coding Questions Bank
  11. Mega Test:
  12. Grand Test:
  13. Live Class Notes: Placement Live Classes
AI summary & chapters

AI Summary

An AI-generated summary of this video lecture.

This lecture segment addresses a problem involving the calculation of the expected number of comparisons for a random search operation within a Binary Search Tree (BST). The instructor presents a specific tree structure and asks students to determine the average search cost. The solution involves identifying the depth of every node, summing these depths to find the total comparisons required for all keys, and then dividing by the total number of nodes to find the mean. The final result is matched against multiple-choice options to identify the correct answer.

Chapters

  1. 0:00 – 1:48 00:00-01:48

    The video opens with the problem statement 'Consider the following Binary Search Tree' alongside a diagram. The tree has root 10, left child 5, right child 20. Node 5 has left child 4. Node 20 has left child 15 and right child 30. Node 4 has a right child 11. The question text appears: 'If we randomly search one of the keys present in above BST, what would be the expected number of comparisons?' Options (A) 2.75, (B) 2.25, (C) 2.57 are visible. The instructor begins solving by annotating the tree with comparison counts. He marks '1' next to 10, '2' next to 5 and 20, '3' next to 4, 15, and 30, and '4' next to 11. He formulates the summation equation on the right side of the screen: 1 + 2 + 2 + 3 + 3 + 3 + 4. He underlines the sum. He divides the sum by 7, writing the fraction 18/7. He performs long division to get 2.57 and circles option (C).

The video effectively teaches how to compute the average search cost in a BST. By assigning a cost equal to the node's depth plus one and averaging these costs across all nodes, the expected number of comparisons is derived. This method highlights that unbalanced trees can have higher average search costs compared to balanced ones.

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