Using the code source screenshot (C++) explain in depth what each line of code does.
Using the code source screenshot (C++) explain in depth what each line of code does.
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
Related questions
Question
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Using the code source screenshot (C++) explain in depth what each line of code does.
![7
#include <iostream>
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using namespace std;
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struct node {
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int info;
struct node *left;
struct node *right;
};
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struct node *generateNode(int value) {
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struct node *temp = (struct node*) malloc(sizeof(struct node));
temp->info = value;
temp->left = temp->right = NULL;
return temp;
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}
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void preorder(struct node *root) {
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if (root != NULL) {
cout << root->info < " ";
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preorder(root->left);
preorder(root->right);
}
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}
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struct node* addNode(struct node* node, int value) {
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if (node == NULL)
return generateNode(value);
if (value < node->info)
node->left = addNode(node->left, value);
else if (value > node->info)
node->right =
return node;
}
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addNode(node->right, value);
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int main() {
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struct node *root = NULL;
root = addNode(root, 6);
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addNode(root, 2);
addNode(root, 1);
addNode(root, 4);
addNode(root, 3);
addNode(root, 5);
addNode(root, 8);
addNode(root, 7);
addNode(root, 9);
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cout << "\nThe pre-order tree traversal of the nodes’ values is: ";
preorder(root);
cout << "\n" << endl;
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return 0;
}
65](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcb9da028-c1e8-4fb4-9df2-7f4287e8030e%2F711ebecb-bb8a-4de4-af76-b37fe171b86d%2Fa1a2slb_processed.png&w=3840&q=75)
Transcribed Image Text:7
#include <iostream>
8
using namespace std;
9.
10
struct node {
11
12
int info;
struct node *left;
struct node *right;
};
13
14
15
16
17
struct node *generateNode(int value) {
18
struct node *temp = (struct node*) malloc(sizeof(struct node));
temp->info = value;
temp->left = temp->right = NULL;
return temp;
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}
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void preorder(struct node *root) {
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if (root != NULL) {
cout << root->info < " ";
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preorder(root->left);
preorder(root->right);
}
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31
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33
}
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struct node* addNode(struct node* node, int value) {
36
if (node == NULL)
return generateNode(value);
if (value < node->info)
node->left = addNode(node->left, value);
else if (value > node->info)
node->right =
return node;
}
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addNode(node->right, value);
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int main() {
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struct node *root = NULL;
root = addNode(root, 6);
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addNode(root, 2);
addNode(root, 1);
addNode(root, 4);
addNode(root, 3);
addNode(root, 5);
addNode(root, 8);
addNode(root, 7);
addNode(root, 9);
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cout << "\nThe pre-order tree traversal of the nodes’ values is: ";
preorder(root);
cout << "\n" << endl;
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return 0;
}
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