Introduction to Java Programming and Data Structures Comprehensive Version (11th Edition)
11th Edition
ISBN: 9780134700144
Author: Liang
Publisher: PEARSON
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Expert Solution & Answer
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.
Expert Solution & Answer
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Check out a sample textbook solutionStudents have asked these similar questions
can u solve this question
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
Chapter 26 Solutions
Introduction to Java Programming and Data Structures Comprehensive Version (11th Edition)
Ch. 26.2 - Prob. 26.2.1CPCh. 26.2 - Prob. 26.2.2CPCh. 26.2 - Prob. 26.2.3CPCh. 26.3 - Prob. 26.3.1CPCh. 26.3 - Prob. 26.3.2CPCh. 26.3 - Prob. 26.3.3CPCh. 26.4 - Prob. 26.4.1CPCh. 26.4 - Prob. 26.4.2CPCh. 26.4 - Prob. 26.4.3CPCh. 26.4 - Prob. 26.4.4CP
Ch. 26.5 - Use Listing 26.2 as a template to describe the...Ch. 26.6 - Prob. 26.6.1CPCh. 26.6 - Prob. 26.6.2CPCh. 26.6 - Prob. 26.6.3CPCh. 26.6 - Prob. 26.6.4CPCh. 26.7 - Prob. 26.7.1CPCh. 26.7 - Prob. 26.7.2CPCh. 26.7 - Prob. 26.7.3CPCh. 26.7 - Prob. 26.7.4CPCh. 26.8 - Prob. 26.8.1CPCh. 26.8 - Prob. 26.8.2CPCh. 26.8 - Prob. 26.8.3CPCh. 26.9 - Prob. 26.9.1CPCh. 26.9 - Prob. 26.9.2CPCh. 26.9 - Prob. 26.9.3CPCh. 26 - Prob. 26.5PE
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