Introduction to Java Programming and Data Structures  Comprehensive Version (11th Edition)
Introduction to Java Programming and Data Structures Comprehensive Version (11th Edition)
11th Edition
ISBN: 9780134700144
Author: Liang
Publisher: PEARSON
Expert Solution & Answer
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Chapter 26.8, Problem 26.8.1CP

Explanation of Solution

AVL tree: It is a self-balancing binary search tree. If the tree is not balanced, the tree performs rotation operation.

Insertion of the nodes 1, 2, 3, 4, 10, 9, 7, 5, 8, 6:

Insertion of 1 and 2:

  • First the value 1 is inserted as the node.
  • The value 2 is greater than 1 and it becomes the right child of 1.

Insertion of 3:

  • Next the value 3 is greater than 1 and it goes to the right of 1.
  • The value 3 is again greater than 3 and it goes to the right of 2.

  • Now the tree gets rebalanced.
  • After inserting 3, the tree becomes,

Insertion of 4:

  • The value 4 is greater than 1 and it goes to the right of 1.
  • The value 4 is greater than 3 and it goes to the right of 3.

Insertion of 10:

  • The value 10 is greater than 1 and it goes to the right of 1.
  • The value 10 is greater than 3 and it goes to the right of 3.
  • The value 10 is greater than 4 and it goes to the right of 4.

  • Now the tree gets rebalanced.
  • After inserting 10, the tree becomes,

Insertion of 9:

  • The value 9 is greater than 2 and it goes to the right of 2.
  • The value 9 is greater than 4 and it goes to the right of 4.
  • The value 9 is lesser than 10 and it goes to the left of 10.

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1. Unsigned Integers If we have an n-digit unsigned numeral dn-1d n-2...do in radix (or base) r, then the value of that numeral is n−1 r² di Σi=0 which is basically saying that instead of a 10's or 100's place we have an r's or r²'s place. For binary, decimal, and hex r equals 2, 10, and 16, respectively. Just a reminder that in order to write down a large number, we typically use the IEC or SI prefixing system: IEC: Ki = 210, Mi = 220, Gi = 230, Ti = 240, Pi = 250, Ei = 260, Zi = 270, Yi = 280; SI: K=103, M = 106, G = 109, T = 10¹², P = 1015, E = 10¹8, Z = 1021, Y = 1024. 1.1 Conversions a. (15 pts) Write the following using IEC prefixes: 213, 223, 251, 272, 226, 244 21323 Ki8 Ki 223 23 Mi 8 Mi b. (15 pts) Write the following using SI prefixes: 107, 10¹7, 10¹¹, 1022, 1026, 1015 107 10¹ M = 10 M = 1017102 P = 100 P c. (10 pts) Write the following with powers of 10: 7 K, 100 E, 21 G 7 K = 7*10³
answer shoul avoid using AI and should be basic and please explain
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