Binary Search Tree shows empty - c++

struct node
{
int data;
node *left,*right;
};
class bst
{
public:
node *root;
bst(){root = NULL;}
void bst_insert(node*,int);
void inorder(node*);
};
void bst::bst_insert(node* x, int d) {
if (x== NULL) {
node* tmp = new node;
tmp->data = d;
tmp->left = NULL;
tmp->right = NULL;
x= tmp;
}
else if (d <= x->data)
bst_insert(x->left,d);
else
bst_insert(x->right,d);
}
void bst::inorder(node* x) {
if(x != NULL) {
inorder(x->left);
cout << x->data << " ";
inorder(x->right);
}
}
int main() {
bst b;
b.bst_insert(b.root,3);
b.bst_insert(b.root,2);
b.inorder(b.root);
}
bst is a class with member node* root (initialize with null on constructor)
Binary Search Tree display in order always shows empty.
What is wrong with the code ?
the code seems fine, but always bst has no value and always show empty, and root is null !!!

No code anywhere sets root to anything other than NULL. When you call inorder, it does nothing since root is NULL.
b.bst_insert(b.root,3);
Since root is NULL at first, this is equivalent to:
b.bst_insert(NULL,3);
This doesn't attach the newly-created node to anything.

Related

I wanted to implement a BST and tried using vector for input

I wanted to implement a BST class with a vector and somehow its not working. I just wanted to know the reason why its not working.
The main reason that I can think of that root in the BST always remain NULL.
I wanted to experiment ways to use classes in data structures.
#include<iostream>
#include<vector>
using namespace std;
class Node{
public:
int data;
Node* left ;
Node* right ;
Node(int val){
data = val;
left = NULL;
right = NULL;
}
};
class BST{
public:
Node* root = NULL;
void insert(Node* r,int data){
Node* new_node = new Node(data);
if(r == NULL){
r = new_node;
}
if(data < r->data){
if(r->left == NULL){
r->left = new_node;
}
else{
insert(r->left,data);
}
}else if(data > r->data){
if(r->right == NULL){
r->right = new_node;
}
else{
insert(r->right,data);
}
}else{
return;
}
return;
}
BST(vector<int> bst_array){
for(int i = 0; i<bst_array.size(); i++){
insert(root,bst_array[i]);
}
}
void print_t(Node* r){
if(r == NULL){
cout<<"NULL";
return;
}
else{
print_t(r->left);
cout<<r->data<<" ";
print_t(r->right);
}
}
};
int main(){
vector<int> v = {1,3,5,44,23,78,21};
BST* tr = new BST(v);
tr->print_t(tr->root);
return 0;
}
There seem to be a logical mistake on my end please help me find it.
Thanks in advance.
The reason is that root is never assigned another value after its initialisation to NULL. Passing root as argument to the insert method can never alter root itself, as it is not the address of root that is passed, but its value.
Some other remarks:
insert always starts by creating a new node, at every step of the recursion. This is a waste of node creation. In the end you just need one new node, so only create it when its position in the tree has been identified.
The final else is not needed, as all it does is execute a return, which it would have done anyway without that else block
As insert is a method of BST, it is a pity that it requires a node as argument. You would really like to just do insert(data) and let it take care of it. For that to happen I suggest to move your insert method to the Node class, where the this node takes over the role of the argument. Then the BST class could get a wrapping insert method that forwards the job to the other insert method.
Instead of NULL use nullptr.
To solve the main issue, there are many solutions possible. But after making the above changes, it is quite easy to assign to root in the simplified insert method on the BST class.
Here is how it could work:
class Node{
public:
int data;
Node* left ;
Node* right ;
Node(int val){
data = val;
left = nullptr;
right = nullptr;
}
void insert(int data) {
if (data < this->data) {
if (this->left == nullptr) {
this->left = new Node(data);
} else {
this->left->insert(data);
}
} else if (data > this->data) {
if (this->right == nullptr) {
this->right = new Node(data);
} else {
this->right->insert(data);
}
}
}
};
class BST {
public:
Node* root = nullptr;
void insert(int data) {
if (root == NULL) { // Assign to root
root = new Node(data);
} else { // Defer the task to the Node class
root->insert(data);
}
}
BST(vector<int> bst_array){
for(int i = 0; i<bst_array.size(); i++){
insert(bst_array[i]); // No node argument
}
}
/* ...other methods ...*/
}

void insert(int ) method for BST tree/C++

I have been trying to get this function working for the longest time now. It is part of an assignment for an online course, but it seems no matter what I submit, the function fails for both the empty child test and the left child test. See code below. The main() function is deliberately commented out. Any info./input is much appreciated.
// C++ binary trees and stuff;
//
#include <iostream>
#include <cstdio>
#include <string>
#include <vector>
using namespace std;
class BST
{
public:
int data;
BST *left;
BST *right;
//BST *root;
// BST() constructor
BST (int num)
{
data = num;
left = nullptr;
right = nullptr;
root = nullptr;
}
// constructors for root node(s), initializing as root when no values exist yet;
BST() : root (nullptr){}
BST (BST *rootNode) : root(rootNode){}
void insert (int value)
{
BST *newNode = new BST();
newNode = root;
if (root == nullptr)
{
root = new BST (value);
}
else
{
root->data = value;
}
// check if newNode's value equals the passed-in value:
if (value == root->data)
{
//cout << "\nWarning! Value already exists in tree, so nothing will be done.\n";
return;
}
// check if value is < or > newNode's value:
if (value <= root->data)
{
if (root->left == nullptr)
{
// make a new node as the left child of this node,
root->left = new BST(value);
}
else
{
// recursively call insert() on tree's left side,
root->left->insert(value);
}
}
else
{
if (root->right == nullptr)
{
// make a new node as the right child of this node,
root->right = new BST(value);
}
else
{
// recursively call insert() on tree's right side,
root->right->insert(value);
}
}
}
public:
BST *root;
};
/*
int main (int argc, char *argv[])
{
//...insert code here,
// create nodes,...
BST rootNode(5);
BST leftNode(4);
BST rightNode(6);
// connect the nodes to the tree via rootNode.left and rootNode.right,..
rootNode.left = &leftNode;
rootNode.right = &rightNode;
printf ("\nData (root) value = %i, rootNode.left = %i, and rootNode.right = %i\n",
rootNode.data, rootNode.left->data, rootNode.right->data);
cout << "\n\nHello, Solar System!\n";
return 0;
}
*/
Okay, here's my suggestion. You need to reformat your code. You need two classes. You need a BST, and you need a Node. The various methods to add/remove/traverse are part of the BST class. Nodes are just Nodes.
So:
class BST_Node {
public:
int value;
BST_Node * left = nullptr;
BST_Node * right = nullptr;
// Define constructors, etc.
};
class BST {
public:
BST_Node * root = nullptr;
BST_Node * insert(int value);
void insertNode(BST_Node *node);
void insertNodeBelow(BST_Node *nodeToInsert, BST_Node *startingNode);
};
BST_Node * BST::insert(int value) {
BST_Node * node = new BST_Node(value);
insertNode(node);
return node;
}
void BST::insertNode(BST_Node *node) {
if (node == nullptr) {
return;
}
if (root == nullptr) {
root = node;
}
else {
insertNodeBelow(node, root);
}
}
void BST::insertNodeBelow(BST_Node *node, BST_Node *startingNode) {
if (node == nullptr || startingNode == nullptr) {
return;
}
if (node->value < startingNode->value) {
if (startingNode->left != nullptr) {
insertNodeBelow(node, startingNode->left);
}
else {
startingNode->left = node;
}
}
else {
if (startingNode->right != nullptr) {
insertNodeBelow(node, startingNode->right);
}
else {
startingNode->right = node;
}
}
}
How this works... First, the logic of how to store nodes is in BST. Nodes don't care. Second, I made methods for either inserting a value or a node. Because I think that's handy. That should be fairly easy to understand.
The root node can be null, if so, then your inserted node is now root. Otherwise it calls the recursive insertion function. Now, you could simplify this a little, but I didn't want to get too clever.
So it's simple. We look to see where it belongs relative to the point we're at (initially the root). Either we go into the left branch or the right branch. But that branch could be empty, so you just plop it right in. If it's not empty, then you recurse.
I didn't test it.

Pointer of struct changes

#include<stack>
#include<iostream>
class Tree{
private:
struct tree{
int val;
tree * lChild;
tree * rChild;
tree * Parent;
};
tree *root;
public:
Tree();
void insert(int x);
};
Tree::Tree(){
root = NULL;
std::cout<<"ROOT inside constructor : "<<root<<std::endl;
}
void Tree::insert(int x){
tree *wst;
wst->val = x;
wst->lChild = NULL;
wst->rChild = NULL;
tree *temp = root;
tree *p = NULL;
std::cout<<"ROOT inside insert : "<<root<<std::endl;
while(temp != NULL){
p = temp;
if(x < temp->val)
temp = temp->lChild;
else
temp = temp->rChild;
}
std::cout<<x<<std::endl;
wst->Parent = p;
if(p == NULL){
root = wst;
}
else{
if(x < p->val)
p->lChild = wst;
else
p->rChild = wst;
}
}
int main(){
Tree tree;
tree.insert(404);
}
I want to check if pointer root is equal to NULL, but it does not seems too work. It seems like the pointer changes from 0 to 0x4 when I am inside the method insert. How can I check if pointer of struct is equal NULL?
EDIT In the insert method if tree doesn't have any nodes it should not enter first while loop, as root should be equall NULL. And my problem is that it enters this loop anyway and crashes when it checks for temp childrens(that are still not defined).
What does wst point to?
tree *wst;
wst->val = x;
wst->lChild = NULL;
wst->rChild = NULL;
// [...]
wst->Parent = p;
Whoops! Your program has undefined behaviour. No wonder it crashes. :)
You probably need tree* wst = new tree(); there. Don't forget to delete your nodes in the Tree destructor, too!
And I'd advise against having a type Tree plus a type tree; perhaps call the latter Node instead?

Counting occurrence in singly linked list by nodes

I am writing a simple app that gets a list and saves the objects as nodes in a singly linked list and we can add(), remove(), copy(), etc. each node depending on the given data set. each node has a char value which is our data and an int count which counts the occurrence of the related char.
e.g. for a list like
a, a, b, b, c, a
there would be three nodes (since there are three different characters) which are:
[a,3,*next] -> [b,2,*next] -> [c,1,*next] -> nullptr
bool isAvailable() checks if the data is already in the list or not.
Q: When inserting a data there are two options:
The data has not been entered: so we have to create a newNodewith the given data, count=1and *next=NULL.
The data is already entered: so we have to count++ the node that has the same data.
I know if the given data is available or not, but how can I point to the node with same data?
Here's the code:
#include "stdafx.h"
#include<iostream>
using namespace std;
class Snode
{
public:
char data;
int count;
Snode *next;
Snode(char d, int c)
{
data = d;
count = c;
next = NULL;
}
};
class set
{
private:
Snode *head;
public:
set()
{
head = NULL;
tail = NULL;
}
~set();
void insert(char value);
bool isAvailable(char value);
};
set::~set()
{
Snode *t = head;
while (t != NULL)
{
head = head->next;
delete t;
}
}
bool set::isAvailable(char value)
{
Snode *floatingNode = new Snode(char d, int c);
while(floatingNode != NULL)
{
return (value == floatingNode);
floatingNode->next = floatingNode;
}
}
void set::insert(char value)
{
Snode *newNode = new Snode(char d, int c);
data = value;
if (head == NULL)
{
newNode->next = NULL;
head = newNode;
newNode->count++;
}
else
{
if(isAvailable)
{
//IDK what should i do here +_+
}
else
{
tail->next= newNode;
newNode->next = NULL;
tail = newNode;
}
}
}
I know if the given data is available or not, but how can I point to the node with same data?
You'll need to start at the head of the list and iterate along the list by following the next pointers until you find the node with the same data value. Once you've done that, you have your pointer to the node with the same data.
Some other notes for you:
bool set::isAvailable(char value)
{
Snode *floatingNode = new Snode(char d, int c);
while(floatingNode != NULL)
{
return (value == floatingNode);
floatingNode->next = floatingNode;
}
}
Why is this function allocating a new Snode? There's no reason for it to do that, just initialize the floatingNode pointer to point to head instead.
This function always returns after looking at only the first node in the linked list -- which is not the behavior you want. Instead, it should return true only if (value == floatingNode); otherwise it should stay inside the while-loop so that it can go on to look at the subsequent nodes as well. Only after it drops out of the while-loop (because floatingNode finally becomes NULL) should it return false.
If you were to modify isAvailable() slightly so that instead of returning true or false, it returned either floatingPointer or NULL, you'd have your mechanism for finding a pointer to the node with the matching data.
e.g.:
// Should return either a pointer to the Snode with data==value,
// or NULL if no such Snode is present in the list
Snode * set::getNodeWithValueOrNullIfNotFound(char value) const
{
[...]
}
void set::insert(char value)
{
Snode * theNode = getNodeWithValueOrNullIfNotFound(value);
if (theNode != NULL)
{
theNode->count++;
}
else
{
[create a new Snode and insert it]
}
}
You had a lot of problems in your code, lets see what are they:
First of all, Snode doesn't need to be a class, rather you can go with a simple strcut; since we need everything public.(not a mistake, but good practice)
You could simple initialize count = 1 and next = nullptr, so that no need of initializing them throw constructor. The only element that need to be initialized through constructor is Snod's data.
Since c++11 you can use keyword nullptr instead of NULL, which denotes the pointer literal.
Member function bool set::isAvailable(char value) will not work as you think. Here you have unnecessarily created a new Snode and cheacking whether it points to nullptr which doesn't allow you to even enter the loop. BTW what you have written in the loop also wrong. What do you mean by return (value == floatingNode); ? floatingNode is a Snode by type; not a char.
Hear is the correct implementation. Since we don't wanna overwrite the head, will create a Node* pointer and assign head to it. Then iterate through list until you find a match. If not found, we will reach the end of the isAvailable() and return false.
inline bool isAvailable(const char& value)
{
Node *findPos = head;
while(findPos != nullptr)
{
if(findPos -> data == value) return true;
else findPos = findPos->next_node;
}
return false;
}
In void set::insert(char value), your logic is correct, but implementation is wrong. Following is the correct implementation.(Hope the comments will help you to understand.
void insert(const char& value)
{
if(head == nullptr) // first case
{
Node *newNode = new Node(value);
newNode->next_node = head;
head = newNode;
}
else if(isAvailable(value)) // if node available
{
Node *temp = head;
while(temp->data != value) // find the node
temp = temp->next_node;
temp->count += 1; // and count it by 1
}
else // all new nodes
{
Node *temp = head;
while(temp->next_node != nullptr) // to find the null point (end of list)
temp = temp->next_node;
temp = temp->next_node = new Node(value); // create a node and assign there
}
}
Your destructor will not delete all what you created. It will be UB, since your are deleting newly created Snode t ( i.e, Snode *t = head;). The correct implementation is as bellow.(un-comment the debugging msg to understand.)
~set()
{
Node* temp = head;
while( temp != nullptr )
{
Node* next = temp->next_node;
//std::cout << "deleting \t" << temp->data << std::endl;
delete temp;
temp = next;
}
head = nullptr;
}
Last but not least, the naming (set) what you have here and what the code exactly doing are both different. This looks more like a simple linked list with no duplicates. This is however okay, in order to play around with pointers and list.
To make the code or iteration more efficient, you could do something like follows. In the isAvailable(), in case of value match/ if you found a node, you could simply increment its count as well. Then in insert(), you can think of, if node is not available part.
Hope this was helpful. See a DEMO
#include <iostream>
// since you wanna have all of Node in public, declare as struct
struct Node
{
char data;
int count = 1;
Node* next_node = nullptr;
Node(const char& a) // create a constrcor which will initilize data
: data(a) {} // at the time of Node creation
};
class set
{
private:
Node *head; // need only head, if it's a simple list
public:
set() :head(nullptr) {} // constructor set it to nullptr
~set()
{
Node* temp = head;
while( temp != nullptr )
{
Node* next = temp->next_node;
//std::cout << "deleting \t" << temp->data << std::endl;
delete temp;
temp = next;
}
head = nullptr;
}
inline bool isAvailable(const char& value)
{
Node *findPos = head;
while(findPos != nullptr)
{
if(findPos -> data == value) return true;
else findPos = findPos->next_node;
}
return false;
}
void insert(const char& value)
{
if(head == nullptr) // first case
{
Node *newNode = new Node(value);
newNode->next_node = head;
head = newNode;
}
else if(isAvailable(value)) // if node available
{
Node *temp = head;
while(temp->data != value) // find the node
temp = temp->next_node;
temp->count += 1; // and count it by 1
}
else // all new nodes
{
Node *temp = head;
while(temp->next_node != nullptr) // to find the null point (end of list)
temp = temp->next_node;
temp = temp->next_node = new Node(value);
}
}
void print() const // just to print
{
Node *temp = head;
while(temp != nullptr)
{
std::cout << temp->data << " " << temp->count << "\n";
temp = temp->next_node;
}
}
};
int main()
{
::set mySet;
mySet.insert('a');
mySet.insert('a');
mySet.insert('b');
mySet.insert('b');
mySet.insert('c');
mySet.insert('a');
mySet.print();
return 0;
}

Write a program, which converts the expression a b + c d e + * * into an expression tree using stack.

description
I don't know how to do this task.... but i just created a tree and enter value..can anyone please help me to do this task...the Stack is also of node type and we have to push value of operators like ab+ so we will push a as node then b as node and when + will come we make a tree and a and b will be its leafs node.
.Code
#include<iostream>
using namespace std;
class Node{
public:
int data;
Node *left;
Node *right;
Node()
{
data = 0;
left = NULL;
right = NULL;
}
};
class Tree
{
Node *root;
void insert(int d, Node *node)
{
if (d < node->data)
{
if (node->left == NULL)
{
Node *leaf = new Node();
leaf->data = d;
node->left = leaf;
}
else
{
insert(d, node->left);
}
}
else
{
if (node->right == NULL)
{
Node *leaf = new Node();
leaf->data = d;
node->right = leaf;
}
else
{
insert(d, node->right);
}
}
}
void inOrderDisplay(Node *subRoot)
{
if (subRoot != NULL)
{
inOrderDisplay(subRoot->left);
cout << subRoot->data << " ";
inOrderDisplay(subRoot->right);
}
}
void postOrderDisplay(Node *subRoot)
{
if (subRoot != NULL)
{
postOrderDisplay(subRoot->left);
postOrderDisplay(subRoot->right);
cout << subRoot->data << " ";
}
}
void preOrderDisplay(Node *subRoot)
{
if (subRoot != NULL)
{
cout << subRoot->data << " ";
preOrderDisplay(subRoot->left);
preOrderDisplay(subRoot->right);
}
}
void deleteSubtree(Node *subRoot)
{
if (subRoot != NULL)
{
deleteSubtree(subRoot->left);
deleteSubtree(subRoot->right);
cout << "\ndeleting: " << subRoot->data;
delete subRoot;
subRoot = NULL;
}
}
public:
Tree()
{
root = NULL;
}
~Tree()
{
deleteAll();
}
void insert(int d)
{
if (root == NULL)
{
Node *leaf = new Node();
leaf->data = d;
root = leaf;
}
else
{
insert(d, root);
}
}
void inOrderDisplay()
{
inOrderDisplay(root);
}
void postOrderDisplay()
{
postOrderDisplay(root);
}
void preOrderDisplay()
{
preOrderDisplay(root);
}
void deleteAll()
{
deleteSubtree(root);
}
};
.Main Class:
#include<iostream>
#include"task1.h"
using namespace std;
void main()
{
Tree tree;
tree.insert(10);
tree.insert(6);
tree.insert(14);
tree.insert(5);
tree.insert(8);
tree.insert(11);
tree.insert(18);
cout << endl;
system("pause");
//tree.deleteAll();
}
Based on the code you have here, you only have a void insert(int d, Node *node) function, no void insert(operator o, Node *node) function.
I think this shows that you missed an important point here. Every node in the tree can either be an integer (as you did) or an operator. In both cases, I'd call it a string. Every node that is not a leaf must be an operator, and all leafs must be integers (or strings that represents operators/integer in our case).
Then, iterating over your input, the first three item should result in something like:
+
/ \
a b
The next step would be to build more sub trees (not sure of the definition of the input you have), keep them in your stack and then construct more inner nodes of the tree.
So if the tree I showed above is called Tree(+) (for ease of use), and the initial stack was [a,b,+,c,d,e,*,*], then after one iteration you'll have [Tree(+),c,d,e,*,*] and you continue from there.