Computer Science Illuminated
7th Edition
ISBN: 9781284155617
Author: Nell Dale, John Lewis
Publisher: Jones & Bartlett Learning
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Question
Chapter 8, Problem 47E
Program Plan Intro
Binary search tree:
- Binary search tree is a tree; the nodes are sorted in the semantic order.
- Binary search tree nodes can have zero, one, or two children.
- Node without children is called a leaf or end node.
- A node that does not have a superior node is called a root node or starting node.
- In binary search tree, the left subtree contains the nodes lesser than the root node.
- The right subtree contains the nodes greater than the root node.
- The binary search will performed until finding a search node or reaching the end of the tree.
Expert Solution & Answer
Explanation of Solution
Insertion of elements in the binary search tree:
Given elements to construct a binary search tree are as follows:
50, 72, 96, 107, 26, 12, 11, 9, 2, 10, 25, 51, 16, 17, 95.
Step 1:
Insert the element 50, which is the root node of a binary search tree.
Step 2:
Next element is 72, which is greater than root node element 50, so insert the element 72 as the right child of the root node.
Step 3:
- Next element is 96, which is greater than root node element 50, so insert the element 96 as the right child of the root node.
- The root node already contains the element 72 as the right child and the element 96 is greater than the element 72.
- So, insert the element 96 as the right child of 72.
Step 4:
- Next element is 107, which is greater than root node element 50, so insert the element 107 as the right child of the root node.
- The root node already contains the element 72 as the right child and the element 107 is greater than the element 72.
- The element 72 already contains the element 96 as the right child and the element 107 is greater than the element 96.
- So, insert the element 107 as the right child of 96.
Step 5:
Next element is 26, which is less than root node value 50, so insert the element 26 as the left child of the root node.
Step 6:
- Next element is 12, which is less than the root node value 50, so insert the element 12 as left child of the root node.
- The root node already contains the element 26 as the left child and the element 12 is less than the element 26.
- So, insert the element 12 as left child of the element 26.
Step 7:
- Next element is 11, which is less than the root node value 50, so insert the element 11 as left child of the root node.
- The root node already contains the element 26 as the left child and the element 11 is less than the element 26.
- The element 26 already contains the element 12 as the left child and the element 11 is lesser than the element 12.
- So, insert the element 11 as left child of the element 12.
Step 8:
- Next element is 9, which is less than the root node value 50, so insert the element 9 as left child of the root node.
- The root node already contains the element 26 as the left child and the element 9 is less than the element 26.
- The element 26 already contains the element 12 as the left child and the element 9 is lesser than the element 12.
- The element 12 already contains the element 11 as the left child and the element 9 is lesser than the element 11.
- So, insert the element 9 as left child of the element 11.
Step 9:
- Next element is 2, which is less than the root node value 50, so insert the element 2 as left child of the root node.
- The root node already contains the element 26 as the left child and the element 2 is less than the element 26.
- The element 26 already contains the element 12 as the left child and the element 2 is lesser than the element 12.
- The element 12 already contains the element 11 as the left child and the element 2 is lesser than the element 11.
- The element 11 already contains the element 9 as the left child and the element 2 is lesser than the element 9.
- So, insert the element 2 as left child of the element 9.
Step 10:
- Next element is 10, which is less than the root node value 50, so insert the element 10 as left child of the root node.
- The root node already contains the element 26 as the left child and the element 10 is less than the element 26.
- The element 26 already contains the element 12 as the left child and the element 10 is lesser than the element 12.
- The element 12 already contains the element 11 as the left child and the element 10 is lesser than the element 11.
- The element 11 already contains the element 9 as the left child and the element 10 is greater than the element 9.
- So, insert the element 10 as right child of the element 9.
Step 11:
- Next element is 25, which is less than the root node value 50, so insert the element 25 as left child of the root node.
- The root node already contains the element 26 as the left child and the element 25 is less than the element 26.
- The element 26 already contains the element 12 as the left child and the element 25 is greater than the element 12.
- So, insert the element 25 as right child of the element 12.
Step 12:
- Next element is 51, which is greater than root node element 50, so insert the element 51 as the right child of the root node.
- The root node already contains the element 72 as the right child and the element 51 is lesser than the element 72.
- So, insert the element 51 as the left child of 72.
Step 13:
- Next element is 16, which is less than the root node value 50, so insert the element 16 as left child of the root node.
- The root node already contains the element 26 as the left child and the element 16 is less than the element 26.
- The element 26 already contains the element 12 as the left child and the element 16 is greater than the element 12.
- The element 12 already contains the element 25 as the right child and the element 16 is lesser than the element 25.
- So, insert the element 16 as left child of the element 25.
Step 14:
- Next element is 17, which is less than the root node value 50, so insert the element 17 as left child of the root node.
- The root node already contains the element 26 as the left child and the element 17 is less than the element 26.
- The element 26 already contains the element 12 as the left child and the element 17 is greater than the element 12.
- The element 12 already contains the element 25 as the right child and the element 17 is lesser than the element 25.
- The element 25 already contains the element 16 as the left child and the element 17 is greater than the element 16.
- So, insert the element 17 as right child of the element 16.
Step 15:
- Next element is 95, which is greater than root node element 50, so insert the element 95 as the right child of the root node.
- The root node already contains the element 72 as the right child and the element 95 is greater than the element 72.
- The element 72 already contains the element 96 as the right child and the element 95 is lesser than the element 96.
- So, insert the element 95 as the left child of 96.
Therefore, the above tree is the final constructed binary search tree for the given elements.
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8.4 Self-Bias Configuration
20. Determine Zi. Zo. and A,, for the network of Fig. 8.73 if gf, = 3000 μS and gos = 50 μs.
21. Determine Z, Zo, and A, for the network of Fig. 8.73 if the 20-uF capacitor is removed and the
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+12 V
3.3 ΚΩ
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FIG. 8.73
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+12 V
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10 ΜΩ
1.1 ΚΩ
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Chapter 8 Solutions
Computer Science Illuminated
Ch. 8 - Prob. 1ECh. 8 - Prob. 2ECh. 8 - Prob. 3ECh. 8 - Prob. 4ECh. 8 - Prob. 5ECh. 8 - Prob. 6ECh. 8 - Prob. 7ECh. 8 - Prob. 8ECh. 8 - Prob. 9ECh. 8 - Prob. 10E
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