binary search tree unsorted array - insert

I have a problem trying to insert elements on my BST given an unsorted array. The output gives me a segmentation fault, but I can't pinpoint the location of my error. I'm pretty sure that something is wrong with my insert, but I'm not sure what. If anyone can give me insight that would be much appreciated.
#include <iostream>
using namespace std;
struct BST{
BST* left;
BST* right;
int data;
};
BST* init(int data){
BST* a = new BST;
a->data = data;
a->right = NULL;
a->left = NULL;
return a;
}
BST* insert(BST* root, int data){
if (root == NULL)
root = new BST;
else{
BST* current = root;
if (current->data >= data)
insert(current->right, data);
else
insert(current->left, data);
}
return 0;
}
void inorderTraversal(BST* root)
{
if(root == NULL)
cout << "No node" << endl;
inorderTraversal(root->left);
cout << root->data << endl;
inorderTraversal(root->right);
}
int main(){
BST* tree = init(5);
int a[8] = {1, 3, 5, 6, 3, 9, 10, 46};
cout << "seg fault" << endl;
for (unsigned int i = 0; i < sizeof(a); i++)
insert(tree, a[i]);
cout << "seg fault here" << endl;
}

Actually, I think the problem is in your for loop. You are looping too far. Try using a.size() instead of sizeof(a). With sizeof(a) you are requesting the size of the array in bytes, which is probably returning 32.

Related

A Scapegoat Tree That Just Won't Balance

So, I'm working on this project for Comp 272, Data Structures and Algorithms, and before anyone asks I have no one to help me. It's an online program through Athabasca University and for some unknown reason they didn't supply me with a tutor for this course, which is a first... So... Yeah. The question is as follows:
"(20 marks) Exercise 8.2. Illustrate what happens when the sequence 1, 5, 2, 4, 3 is added to an empty ScapegoatTree, and show where the credits described in the proof of Lemma 8.3 go, and how they are used during this sequence of additions."
This is my code, its complete and it compiles:
/*
Name: Westcott.
Assignment: 2, Question 3.
Date: 08-26-2022.
"(20 marks) Exercise 8.2. Illustrate what happens when the sequence 1, 5, 2, 4, 3 is added to an empty
ScapegoatTree, and show where the credits described in the proof of Lemma 8.3 go, and how they are used
during this sequence of additions."
*/
#include <iostream>
using namespace std;
class Node { // Originally I did this with Node as a subclass of sgTree but I found that this
public: // way was easier. This is actually my second attempt, from scratch, at doing this
int data; // problem. First version developed so many bugs I couldn't keep up with them.
Node* left;
Node* right;
Node* parent;
Node() : data(0), parent(NULL), left(NULL), right(NULL) {};
Node(int x) : data(x), parent(NULL), left(NULL), right(NULL) {};
~Node() {}; // Normally I would do a little more work on clean up but... Yea this problem didn't leave me much room.
Node* binarySearch(Node* root, int x); // The Node class only holds binarySearch in addition to its
// constructors/destructor, and of course the Node*'s left, right and parent.
};
class sgTree { // The sgTree keeps track of the root, n (the number of nodes in the tree), and q which is
public: // as Pat put it a 'high water mark'.
Node* root;
int n;
int q;
sgTree() : root(new Node()), n(1), q(1) {}
sgTree(int x) : root(new Node(x)), n(0), q(0) {}
~sgTree() {
delete root;
}
bool add(int x); // The add function is compounded, within it are findDepth and rebuild.
bool removeX(int x); // removeX works, but it didn't have a big part to play in this question,
int findDepth(Node* addedNode); // but I'll include it to maintain our sorted set interface.
void printTree(Node* u, int space) { // This was extra function I wrote to help me problem solve.
cout << "BINARY TREE DISPLAY" << endl; // this version only prints a title and then it calls printTreeSub on line 46.
cout << "________________________________________________\n\n" << endl;
printTreeSub(u, space);
cout << "________________________________________________\n\n" << endl;
}
int printTreeSub(Node* u, int space); // Function definition for this is on line 81.
int storeInArray(Node* ptr, Node* arr[], int i);// this is our function for storing all the elements of a tree in an array.
int size(Node* u); // this is size, defined on line 74.
void rebuild(Node* u); // And rebuild and buildBalanced are the stars of the show, defined on lines 262 and 282
Node* buildBalanced(Node** a, int i, int ns); // just above the main() funciton.
};
int log32(int q) { // As you can see there's two versions of this function.
int c = 0; // this is supposed to return the log of n to base 3/2.
while (q != 0) { // The version below I got from this website:
q = q / 2; // https://www.geeksforgeeks.org/scapegoat-tree-set-1-introduction-insertion/
c++; // It works fine but I prefer the one I wrote.
} // this is a much simpler function. It just divides q until its zero
return c; // and increments c on each division. Its not exact but it is based on what Pat said
} // in this lecture: https://www.youtube.com/watch?v=OGNUoDPVRCc&t=4852s
/*
static int const log32(int n)
{
double const log23 = 2.4663034623764317;
return (int)ceil(log23 * log(n));
}
*/
int sgTree::size(Node* u) {
if (u == NULL) {
return 0;
}
return 1 + size(u->left) + size(u->right); // Recursion in size();
}
int sgTree::printTreeSub(Node* u, int space) { // Here is my strange print function
if (u == NULL) return space; // I say strange because I'm not even 100% sure
space--; // how I got it to work. The order itself I worked out, but I built it
space -= printTreeSub(u->left, space); // and, originally, got a half decent tree, but then I just kept playing
if (u->right == NULL && u->left == NULL) { // around with increments, decrements, and returned values
cout << "\n\n\n" << u->data << "\n\n\n" << endl; // of space until it just sort of came together.
return 1; // Basically it prints the left most Node first and then prints every node
} // beneath that using recursion. I realized that by setting the for loop
for (int i = space; i >= 0; i--) { // on line 89 I could imitate different nodes having different heights in
cout << " "; // the tree. I figured that using n as an input I could take advantage of
} // the recursion to get an accurate tree. That much I understand.
cout << " " << u->data << "'s children are: "; // But it didn't work out quite how I wanted it to so I just kept playing
if (u->left != NULL) { // with space increments and decrements on different sides of the tree until
cout << u->left->data; // I got something pretty good.
}
else {
cout << "NULL";
}
if (u->right != NULL) {
cout << " and " << u->right->data;
}
else {
cout << " NULL";
}
cout << "\n\n" << endl;
space--;
space -= printTreeSub(u->right, space);
return 1;
}
int sgTree::storeInArray(Node* ptr, Node* a[], int i) { // This function took me a while to figure out.
if (ptr == NULL) { // The recursive insertions of values using i, when
return i; // i is defined by the very same recursion, makes this
} // a bit of a challenge to get your head around.
i = storeInArray(ptr->left, a, i); // Basically its just taking advantage on an inOrder
a[i] = ptr; // transversal to get the values stored into the array
i++; // in order from least to greatest.
return storeInArray(ptr->right, a, i);
}
Node* Node::binarySearch(Node* root, int x) { // I covered this in another question.
if (root->data == x) {
return root;
}
else if (x < root->data) {
if (root->left == NULL) {
return root;
}
return binarySearch(root->left, x);
}
else if (x > root->data) {
if (root->right == NULL) {
return root;
}
return binarySearch(root->right, x);
}
}
bool sgTree::add(int x) { // The add function itself isn't too difficult.
Node* addedNode = new Node(x); // We make a Node using our data, then we search for that Node
Node* parent = root->binarySearch(root, x); // in the tree. I amended binarySearch to return the parent
addedNode->parent = parent; // if it hits a NULL child, on lines 127 and 133.
if (x < parent->data) { // That way the new Node can just go into the returned parents child
parent->left = addedNode; // here is where we choose whether it enters the left or the right.
}
else if (x > parent->data) {
parent->right = addedNode;
}
int h = findDepth(addedNode); // We run findDepth() on the addedNode. I realize that this probably should
// have been a part of the binarySearch, it means we go down
if (h > log32(q)) { // the tree twice instead of once. I did look at changing binarySearch into searchAndDepth
// having binarySearch return an int for the height isn't a problem, but then that would
// mess up removeX and, I don't know. What's more important?
Node* w = addedNode->parent; // If this were going to be a database hosting millions of pieces of data I would give
while (3 * size(w) < 2 * size(w->parent)) { // that alot more consideration but, this is just an exercise after all so...
w = w->parent; // From there, we compare our height to the value output by log32(q) on line 152.
}
rebuild(w); // This expression 3 * size(w) < 2 * size(w->parent) is the formula on page 178 rewritten
//rebuild(root); // as a cross multiplication, clever. It keeps going up the tree until we find the scapegoat w.
// This is a much nicer result.
//See line 311.
} // Now, this is where my problems began. Pat says that this line should read: rebuild(w->parent);
n++; // but when I do that I get an error when w is the root. Because then w->parent is NULL. And in that case
q++; // line 258 throws an error because we're trying to set p equal to NULL's parent. It's not there.
return true; // So my work around was to just offset this by one and send rebuild(w). But that doesn't seem
} // to balance the tree just right. In fact, the best tree results when we replace w with root.
// and just rebalance the whole tree. But in any case, we increment n and q and lets pick this up on line 256.
int sgTree::findDepth(Node* addedNode) {
int d = 0;
while (addedNode != root) {
addedNode = addedNode->parent;
d++;
}
return d;
}
bool sgTree::removeX(int x) {
Node* u = root->binarySearch(root, x);
if (u->left == NULL && u->right == NULL) {
if (u == u->parent->left) {
u->parent->left = NULL;
}
if (u == u->parent->right) {
u->parent->right = NULL;
}
cout << u->data << " deleted" << endl;
n--;
delete u;
return true;
}
if (u->left != NULL && u->right == NULL) {
if (u->parent->left = u) {
u->parent->left = u->left;
}
else if (u->parent->right = u) {
u->parent->right = u->left;
}
cout << u->data << " deleted" << endl;
n--;
delete u;
return true;
}
if (u->left == NULL && u->right != NULL) {
if (u == u->parent->left) {
u->parent->left = u->right;
u->right->parent = u->parent;
}
else if (u == u->parent->right) {
u->parent->right = u->right;
u->right->parent = u->parent;
}
cout << u->data << " deleted" << endl;
n--;
delete u;
return true;
}
if (u->left != NULL && u->right != NULL) {
Node* X = u->right;
if (X->left == NULL) {
X->left = u->left;
if (u->parent != NULL) {
if (u->parent->right == u) {
u->parent->right == X;
}
else if (u->parent->left == u) {
u->parent->left = X;
}
}
else {
root = X;
}
X->parent = u->parent;
cout << u->data << " deleted" << endl;
n--;
delete u;
return true;
}
while (X->left != NULL) {
X = X->left;
}
X->parent->left = NULL;
X->left = u->left;
X->right = u->right;
if (u->parent != NULL) {
X->parent = u->parent;
}
cout << u->data << " deleted" << endl;
n--;
root = X;
delete u;
return true;
}
}
void sgTree::rebuild(Node* u) {
int ns = size(u); // Everything is pretty kosher here. Just get the number of nodes in the subtree.
Node* p = u->parent; // Originally I had n here instead of ns and... I don't want to talk about how long it took me to find that mistake...
/* It's funny because while writing the comments for this I'm like "Oh, hang on, if I just push the definition of p behind the if statement on line 262
and evaluate for whether or not u is NULL instead of p, that should solve all my problems! Yea, no, it doesn't. Because then for some reason it tries rebalancing
empty tree and... Yea I just have to stop myself from trying to fix this because everytime I do I get caught in an infinite loop of me chasing my tail in errors.
I think a solution could be found in buildBalanced, and I literally went through that function line by line, trying to comprehend a work around. I've included at
a photograph of that white board. Yea this is the code that Pat gave us... and its garbage. It doesn't work. Maybe its a C++ thing, I don't know... But I'm
getting frustrated again so I'm going to stop thinking about this part RIGHT HERE, and move on LOL*/
Node** a = new Node * [ns]; // a Node pointer-pointer array... again, another fine piece of code from the textbook. Sorry, trying to stay positive here.
storeInArray(u, a, 0); // See Line 112
if (p == NULL) { // Okay, once we have our array we use buildBalanced to rebuild the subtree with respect to which
root = buildBalanced(a, 0, ns); // child u is relative to its parent.
root->parent = NULL; // See line 281 for buildBalanced().
}
else if (p->right == u) {
p->right = buildBalanced(a, 0, ns);
p->right->parent = p;
}
else {
p->left = buildBalanced(a, 0, ns);
p->left->parent = p;
}
}
Node* sgTree::buildBalanced(Node** a, int i, int ns) { // This is without a doubt one of the hardest functions I've ever had
if (ns == 0) { // the displeasure of trying to understand... Trying to stay positive.
return NULL; // I've gone through it, in a line by line implementation of the array:
} // a[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} you can find that analysis in
int m = ns / 2; // the photo buildBalanced_Analysis.
a[i + m]->left = buildBalanced(a, i, m); // As confusing as it is, I have to admit that it is a beautiful function.
if (a[i + m]->left != NULL) { // It basically uses the two integers i and m to simultaneously
a[i + m]->left->parent = a[i + m]; // regulate the organization of the new tree and to specifically
} // grab the right value from the array when its needed.
a[i + m]->right = buildBalanced(a, i + m + 1, ns - m - 1); // but trying to map this out didn't help me to solve the issues I've been having.
if (a[i + m]->right != NULL) {
a[i + m]->right->parent = a[i + m];
}
return a[i + m];
}
int main() {
sgTree newTree(1);
int a[] = { 5, 2, 4, 3 };
for (int i = 0; i < (sizeof(a) / sizeof(a[0])); i++) {
newTree.add(a[i]);
}
newTree.printTree(newTree.root, newTree.n);
/*
This is a nice test, when paired with rebuild(root), that too me is the only thing that approaches redeeming this whole question.
sgTree newTreeB(1);
int b[] = { 2, 3, 4, 5, 6, 7, 8, 9, 10 };
for (int i = 0; i < (sizeof(b) / sizeof(b[0])); i++) {
newTreeB.add(b[i]);
}
newTreeB.printTree(newTreeB.root, newTreeB.n);
*/
}
Now the issue itself is not that hard to understand. My tree should look like this:
But instead, it looks like this, with 5 at the root and the values 1 and 4 as the leaves:
I'm confident that the problem lives somewhere around line 159 and in those first few calls to buildBalanced. The comments in the code itself elaborate more on the issue. I've spent days just pouring over this trying everything I can think of to make it work and... Yeah... I just can't figure it out.

C++, Pointers, Merging of two sorted linked lists

I took reference of code from GeeksForGeeks for merging of two sorted linked lists.
#include <bits/stdc++.h>
using namespace std;
/* Link list node */
struct Node
{
int data;
struct Node* next;
};
void MoveNode(Node **destRef, Node **sourceRef){
cout << "pp: " << (*sourceRef) << endl;
Node *tempNode = *sourceRef;
cout << "tt:" << tempNode->next << endl;
*sourceRef = tempNode->next;
tempNode->next = *destRef;
*destRef = tempNode;
cout << "qq: " << (*sourceRef) << endl;
}
struct Node* SortedMerge(Node *a, Node *b){
Node *dummy;
Node *tail;
dummy->next = NULL;
tail = dummy;
cout << "hii" << endl;
while(true){
if(a==NULL){
cout << "aa" << endl;
tail->next = b;
break;
}
else if(b==NULL){
cout << "bb" << endl;
tail->next = a;
break;
}
if(a->data < b->data){
cout << "cc" << endl;
MoveNode(&(tail->next), &a);
tail = tail->next;
}
else if(a->data >= b->data){
cout << "dd" << endl;
// cout << "b->data: " << b << endl;
MoveNode(&(tail->next), &b);
// b = b->next;
// cout << "b->data: " << b << endl;
// cout << "b->data: " << b->data << endl;
tail = tail->next;
}
}
return dummy->next;
}
/* Function to insert a node at the beginning of the
linked list */
void push(struct Node** head_ref, int new_data)
{
/* allocate node */
struct Node* new_node =
(struct Node*) malloc(sizeof(struct Node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Function to print nodes in a given linked list */
void printList(struct Node *node)
{
while (node!=NULL)
{
printf("%d ", node->data);
node = node->next;
}
cout << endl;
}
/* Drier program to test above functions*/
int main()
{
/* Start with the empty list */
struct Node* res = NULL;
struct Node* a = NULL;
struct Node* b = NULL;
/* Let us create two sorted linked lists to test
the functions
Created lists, a: 5->10->15, b: 2->3->20 */
push(&a, 15);
push(&a, 10);
push(&a, 5);
push(&b, 20);
push(&b, 3);
push(&b, 2);
/* Remove duplicates from linked list */
res = SortedMerge(a, b);
printf("Merged Linked List is: \n");
printList(res);
return 0;
}
For the given example in the main function, the program gives wrong output if I do not print the node values in the second else if of while loop in SortedMerge function. And if I do print them the program gives correct output. I find it very strange. Can someone please help me out?

How to iterate a string using while loop in C++?

number = 100010001111111
for (int i=0; number.length(); i++) {
while number[i] == 1 {
k++;
}
}
I would like to implement a while-loop as a replacement for the for-loop as shown above.
How could I convert this to a while-loop?
Here's a solution for the problem you mentioned in your comment (Problem - 96A)
#include <iostream>
using namespace std;
int main()
{
cout << "Please enter your players situation" << endl;
std::string str;
cin >> str;
std::string::size_type i = 0;
int NumbersofAppearances = 0;
int ConsectiveNumberSequence = 7; //You can change that to whatever sequence you like
bool IsDangerous=false;
while (i < str.size())
{
if(str[i]=='1' )
{
++NumbersofAppearances;
//We need to check if we reached the consecutive number or not and save it on a different bool variable
if(NumbersofAppearances>=ConsectiveNumberSequence)
IsDangerous=true;
}
else
{
NumbersofAppearances=0;
}
++i;
}
//print out the end result
if (IsDangerous)
cout <<"YES , this is dangerous"<< endl;
else
cout <<"No, this is not dangerous"<< endl;
return 0;
}
And here's a link to Coding ground

Heapsort implementation

enter code here
# include <iostream>
# include <stdlib.h>
# define MAX 10
void heapsort(int A[]);
void Build_MAX_Heap(int A[]);
void MAX_Heapify(int A[],int i);
int Left(int i);
int Right(int i);
void swap(int *num,int *num2);
using namespace std;
int main()
{
int H[100],i;
for(i=0;i<MAX;i++)
H[i]=rand();
cout << "the given array is::" << " ";
for(i=0;i<MAX;i++)
cout << H[i] << "\n";
cout << "\n" << "\n";
heapsort(H);
cout << "the sorted array is ::" << " ";
for(i=0;i<MAX;i++)
cout << H[i] << "\n";
}
void heapsort(int A[])
{
int i,heapsize;
Build_MAX_Heap(A);
for(i=MAX-1;i>0;i--)
{
swap(&A[0],&A[i]);
heapsize=heapsize-1;
MAX_Heapify(A,0);
}
}
void Build_MAX_Heap(int A[])
{
int heapsize,i;
heapsize=MAX;
for(i=(MAX)/2;i>0;i--)
{
MAX_Heapify(A,i);
}
}
void MAX_Heapify(int A[],int i)
{
int l,r,largest,heapsize;
l=Left(i);
r=Right(i);
if(l<=heapsize && A[l]>A[i])
largest=l;
else
largest=i;
if(r<=heapsize && A[r]>A[i])
largest=r;
if(largest!=i)
{
swap(&A[i],&A[largest]);
MAX_Heapify(A,largest);
}
}
int Left(int i)
{
return (2*i);
}
int Right(int i)
{
return (2*i+1);
}
` void swap(int *num1,int *num2)
{
int temp;
temp=*num1;
*num1=*num2;
*num2=temp;
}
whats wrong in my code.its not sorting.It shows the outout but not in the sorted order.please help.thanks for the same
There are a few flaws in your code.
The notation of left child being (2 * i), and right child being (2 * i + 1), is valid, if you are using your array as 1-indexed. But as you are using your arrays as 0-indexed, you must change them to, left child = (2 * i + 1), right child = (2 * i + 2).
Next thing, the variable 'heapsize' in your function 'MAX_Heapify()' is not initialised. You are using an unassigned variable, which is incorrect.
In your heap sort procedure, you are deleting the max element and decreasing the size of the heap. But your are not using the new size anywhere. You are supposed to pass it to 'MAX_Heapify()' so that the procedure knows its bounds.
Lastly, I think you must run the loop for heapify in Build_MAX_Heap till (i >= 0), i.e., for all the elements above (MAX / 2), including the root.
Correct these and re-write your code and I think you should be fine. If you to learn more about Binary Heaps with sketches and code, you can check my blog post on Binary Heaps.
I hope my answer has helped you, if it did, let me know...! ☺

find chars/string in string from vector c++

I have a vector of strings, and I want to count all 'Ace' in the vector. Right now I can only find one...
int main()
{
std::vector<string> vec;
vec.push_back("Ace of Spades");
vec.push_back("Ace");
string value = "Ace";
int cnt = 0;
auto iter = find_if(begin(vec), end(vec), [&](const string &str)
{
return str.find(value) != str.npos;
});
if(iter == end(vec))
cout << "no found" << endl;
else
{
cout << *iter << endl;
cnt++;
cout << cnt++ << endl;
}
}
You could use std::count_if:
auto cnt = count_if(begin(vec),
end(vec),
[&](const string& str) {
return str.find(value) != std::string::npos;
});
Note that this only counts the number of strings containing "Ace", not the total number of occurrences of "Ace" in the vector's elements.
If you just want to count the number of matching elements, you could use std::count_if.
If you also need to do something with them, it would probably be best to forget about the standard library algorithms and use a ranged for like so:
int count = 0;
for (const auto& element : vec) {
if (element.find(value) != std::string::npos) {
std::cout << element << std::endl;
++count;
}
}
std::cout << count << std::endl;

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