Chapter 4, Problem 12C

Explanation of Solution

Finding minimum and maximum of “n” numbers:

• Assume that a binary tree is created with “n” number of nodes.
• A binary tree can consists of fewer left children nodes and fewer right children nodes.
• Sometimes, it may not contain any children node.
• To find the minimum and maximum value, “$\frac{n}{2}$” number of comparisons is required if the total number of children nodes is even.
• “$\frac{\left(\text{n}-\text{1}\right)}{\text{2}}\text{+1}$” number of comparisons is required to find the minimum and maximum value, if the total number of children nodes is odd.
• Hence, there is “m” number of comparisons in worst case.
• Therefore, it can have,

$\text{m+}\frac{\text{m}}{\text{2}}\text{+}\frac{\text{m}}{\text{4}}\text{+}\frac{\text{m}}{\text{8}}\text{+}\mathrm{....}\text{+2=}\text{2m}-\text{2}$

• When the total number of children node is even, total number of comparisons to find the minimum and maximum value is,

$\frac{\text{n}}{\text{2}}\text{+n}-\text{2<}\frac{\text{3n}}{\text{2}}$

• When