BST Practice Question

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

1. 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. 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.

3. 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.