I have to create binary tree in which nodes store a char value. The task is to find the largest lexicographically root to leaf path created by these chars.
The given input should be a string where first char is value to store and after space there are hints in which node it is stored. L means left node, R of course right one.
The output should be a found string and number of chars given in input (not whitespaces).
This is my code. I'm pretty sure that mistake is in rootToLeafPath(), because I've already checked method which creates tree. I'm giving you also, print method if you want to see all paths.
#include <stdio.h>
#include <iostream>
#include <string.h>
int allCharCounter = 0;
char oldest_word[65];
struct Tree_node{
struct Tree_node *right;
struct Tree_node *left;
char value;
};
Tree_node* getTree(struct Tree_node* root){
char c = getchar();
char edge = c;
Tree_node* korzen = new Tree_node();
if(root == NULL){
root = new Tree_node();
}
korzen = root;
while(!feof(stdin)){
c = getchar();
if(c == 82){//R
allCharCounter++;
if(root->right == NULL){
root->right = new Tree_node();
}
root = root->right;
}else if(c == 76){//L
allCharCounter++;
if(root->left == NULL){
root->left = new Tree_node();
}
root = root->left;
}else if(c > 96 && c < 123){
allCharCounter++;
root->value = edge;
root = korzen;
edge = c;
}
}
root->value = edge;
root = korzen;
return root;
}
void printPath(char *path, int length){
int i;
for(i = 0; i <= length; i++){
printf("%c ", path[i]);
}
printf("\n");
}
void rootToLeafPath(struct Tree_node *nodeptr, char *current_path, int index){
if(nodeptr != NULL){
current_path[index] = nodeptr->value;
if(nodeptr->left == NULL && nodeptr->right == NULL){
if(strcmp(oldest_word,current_path)< 0){
//memset(oldest_word,0,sizeof(oldest_word));
strncpy(oldest_word, current_path, 65);
}
//printPath(current_path, index);
}
rootToLeafPath(nodeptr->left, current_path,index+1);
rootToLeafPath(nodeptr->right, current_path,index+1);
}
}
int main(){
struct Tree_node* root = NULL;
struct Tree_node* test = NULL;
root = getTree(root);
char current_path [65] ={};
rootToLeafPath(root, current_path,0);
std::cout<< oldest_word;
fprintf(stdout,"\n%d", allCharCounter+1); //-> ok
}
So for input:
s LR
z LRR
m RR
p LRLRL
k
w LRL
a LL
t L
h R
j LRLR
a LRRR
The output should be:
ktsza
38
But my code creates:
ktszap
38
I thought maybe I need to clear oldest_word before giving it a new value, but didn't work. For me it looks like it remembers longer value which was before. In this example, 'ktswjp' was the word in array before, but then it found new one which was 'ktsza', but the 'p' stayed.
Appreciate any help.
In rootToLeafPath, you assign a value to current_path[index] = nodeptr->value; to store the next character. When you're done with that character, you don't clear it out so it stays in the buffer, resulting in it appearing at the end of strings that should be shorter.
The solution is to reset it to the nul character before you return, with
current_path[index] = '\0';
after your recursive calls to rootToLeafPath are done.
Related
Hi everyone this is my first time in Stackoverflow. I have a question regarding counting the occurrence of words in text file using C++. This is my code so far. I have to create an array struct of index of the word and the counter of each word then store all of them in an AVL tree. After opening the file and read a word, I look for it in the avl tree or trie. If it is there, use the node's index to increment the word's Cnt. If it is not there, add it to the word array and put its position in the next struct and put the structs position in the avl tree. Also I set the struct Cnt to 1. The problem I am having now is it seems like my program doesn't process the counting properly therefore it only prints out 0. Please give me recommendation on how I can fix the bug. Please find my code below:
#include <iostream>
#include <fstream>
#include <string>
#include <cstdlib>
#include <cstring>
#include <ctype.h>
#include <stdio.h>
#include <string>
#include <cctype>
#include <stdlib.h>
#include <stdbool.h>
using namespace std;
struct Node* insert(struct Node* node, int key) ;
void preOrder(struct Node *root) ;
void removePunct(char str[]);
int compareWord(char word1[], char word2[] );
struct Stats {
int wordPos, wordCnt;
};
Stats record[50000];
int indexRec = 0;
char word[50000*10] ;
int indexWord = 0;
int main() {
ifstream fin;
string fname;
char line[200], wordArray[500000];
cout << "Enter the text file name:" << endl;
cin >> fname;
fin.open(fname.c_str());
if (!fin) {
cerr << "Unable to open file" << endl;
exit(1);
}
struct Node *root = NULL;
while (!fin.eof() && fin >> line) { //use getline
for(int n=0,m=0; m!=strlen(line); m+=n) {
sscanf(&line[m],"%s%n",word,&n);
removePunct(word);
//strcpy(&wordArray[indexWord],word);
int flag = compareWord(wordArray, word);
if(flag==-1) {
strcpy(&wordArray[indexWord],word);
record[indexRec].wordPos = indexWord;
record[indexRec].wordCnt = 1;
root = insert(root, record[indexRec].wordPos);
indexWord+=strlen(word)+1;
// indexes of the word array
indexRec++;
cout << wordArray[indexWord] << " ";
} else
record[flag].wordCnt++;
cout << record[indexRec].wordCnt;
cout << endl;
}
/*for(int x = 0; x <= i; x++)
{
cout << record[x].wordPos << record[x].wordCnt << endl;
}*/
}
fin.close();
return 0;
}
void removePunct(char str[]) {
char *p;
int bad = 0;
int cur = 0;
while (str[cur] != '\0') {
if (bad < cur && !ispunct(str[cur]) && !isspace(str[cur])) {
str[bad] = str[cur];
}
if (ispunct(str[cur]) || isspace(str[cur])) {
cur++;
} else {
cur++;
bad++;
}
}
str[bad] = '\0';
for (p= str; *p!= '\0'; ++p) {
*p= tolower(*p);
}
return;
}
int compareWord(char word1[], char word2[] ) {
int x = strcmp(word1, word2);
if (x == 0 ) return x++;
if (x != 0) return -1;
}
struct Node {
int key;
struct Node *left;
struct Node *right;
int height;
};
// A utility function to get maximum of two integers
int max(int a, int b);
// A utility function to get height of the tree
int height(struct Node *N) {
if (N == NULL)
return 0;
return N->height;
}
// A utility function to get maximum of two integers
int max(int a, int b) {
return (a > b)? a : b;
}
/* Helper function that allocates a new node with the given key and
NULL left and right pointers. */
struct Node* newNode(int key) {
struct Node* node = (struct Node*)
malloc(sizeof(struct Node));
node->key = key;
node->left = NULL;
node->right = NULL;
node->height = 1; // new node is initially added at leaf
return(node);
}
// A utility function to right rotate subtree rooted with y
// See the diagram given above.
struct Node *rightRotate(struct Node *y) {
struct Node *x = y->left;
struct Node *T2 = x->right;
// Perform rotation
x->right = y;
y->left = T2;
// Update heights
y->height = max(height(y->left), height(y->right))+1;
x->height = max(height(x->left), height(x->right))+1;
// Return new root
return x;
}
// A utility function to left rotate subtree rooted with x
// See the diagram given above.
struct Node *leftRotate(struct Node *x) {
struct Node *y = x->right;
struct Node *T2 = y->left;
// Perform rotation
y->left = x;
x->right = T2;
// Update heights
x->height = max(height(x->left), height(x->right))+1;
y->height = max(height(y->left), height(y->right))+1;
// Return new root
return y;
}
// Get Balance factor of node N
int getBalance(struct Node *N) {
if (N == NULL)
return 0;
return height(N->left) - height(N->right);
}
// Recursive function to insert key in subtree rooted
// with node and returns new root of subtree.
struct Node* insert(struct Node* node, int key) {
/* 1. Perform the normal BST insertion */
if (node == NULL)
return(newNode(key));
if (key < node->key)
node->left = insert(node->left, key);
else if (key > node->key)
node->right = insert(node->right, key);
else // Equal keys are not allowed in BST
return node;
/* 2. Update height of this ancestor node */
node->height = 1 + max(height(node->left),
height(node->right));
/* 3. Get the balance factor of this ancestor
node to check whether this node became
unbalanced */
int balance = getBalance(node);
// If this node becomes unbalanced, then
// there are 4 cases
// Left Left Case
if (balance > 1 && key < node->left->key)
return rightRotate(node);
// Right Right Case
if (balance < -1 && key > node->right->key)
return leftRotate(node);
// Left Right Case
if (balance > 1 && key > node->left->key) {
node->left = leftRotate(node->left);
return rightRotate(node);
}
// Right Left Case
if (balance < -1 && key < node->right->key) {
node->right = rightRotate(node->right);
return leftRotate(node);
}
/* return the (unchanged) node pointer */
return node;
}
void preOrder(struct Node *root) {
if(root != NULL) {
printf("%d ", root->key);
preOrder(root->left);
preOrder(root->right);
}
}
One problem (I cannot see if this is the only problem) is that you have code like this, deleting all the intermediate lines:
record[indexRec].wordCnt = 1;
if find word fails
indexRec++;
cout << record[indexRec].wordCnt;
So when you have a new word (if I understand the code correctly!) you are printing out the next record. One fix would be:
if (flag==-1)
cout << record[indexRec-1].wordCnt;
else
cout << record[indexRec].wordCnt;
There's a lot of other issues, like compareWord() is very wrong, you should decide if you really want to use C++ or just C with std::cout, the file reading code is odd, you're including both C and C++ versions of standard headers, etc, but these are issues for another question!
I'm trying to build a trie in c++ by implementing it through a B-Tree. Every single node has a key, an array of pointers to other nodes, and a boolean value determining if it's a leaf or not. By default, all pointers point to null when initializing a new node and whenever a new node is added, the previous node points to the new node.
But for some reason, my else if (currentNode.children[arrayPointer] != NULL) is never being called. I'm completely stumped. I'm not sure if it's because I'm labeling something wrong or if somethings not in the right order.
#include <iostream>
using namespace std;
numberOfWords = 0;
numberOfNodes = 0;
struct Node;
struct Node;
struct Node {
char key;
bool isLeaf;
struct Node *children[36]; //26 letters + 10 digits
};
main() {
//initializing the root
Node root;
root.key ='$';
int i = 0;
while (i < 36) {
root.children[i] = NULL;
i++;
}
root.isLeaf = 1;
numberOfNodes++;
insertWord("the", root);
insertWord("and", root);
insertWord("there", root);
cout << numberOfWords << " words found." << endl;
cout << numberOfNodes << " nodes found." << endl;
}
void insertWord(string word, Node currentNode) {
int wordLength = word.length();
char letterToInsert;
letterToInsert = word[0];
int arrayPointer;
arrayPointer = charToInteger(letterToInsert);
//if the node is not found, a new path is inserted
if (currentNode.children[arrayPointer] == NULL) {
insertNewPath(word, currentNode);
}
//if the node with the current letter is found, I want to recursively call
//insertWord and pass in the updated word and the next node.
//For some reason, this statement is never being called.
else if (currentNode.children[arrayPointer] != NULL) {
string updatedWord;
updatedWord = word.substr(1,wordLength);
Node nextNode;
currentNode.children[arrayPointer] = &nextNode;
insertWord(updatedWord, nextNode);
}
}
//This function will keep recursively calling itself and passing in the new word
//and the new node until the each letter is added.
void insertNewPath(string word, Node currentNode) {
int wordLength;
wordLength = word.length();
if (wordLength > 0) {
char letterToInsert;
letterToInsert = word[0];
int arrayPointer;
//charToInteger takes a character and converts it into an integer value - a=0,b=1,...,8=34,9=35
arrayPointer = charToInteger(letterToInsert);
Node newNode;
numberOfNodes++;
newNode.key = letterToInsert;
// Here, I set the current node's pointer to point to the new node.
// This should cause line 168 to execute when entering an existing
// letter to the trie but for some reason it never does.
currentNode.children[arrayPointer] = &newNode;
//setting every pointer in the new node to null
int i = 0;
while (i < 36) {
newNode.children[i] = NULL;
i++;
}
currentNode.isLeaf = 0;
string updatedWord;
updatedWord = word.substr(1, wordLength);
insertNewPath(updatedWord, newNode);
}
else if (wordLength == 0){
currentNode.isLeaf = 1;
//Once all the letters have been added, the last
//node/letter to be added is set as a leaf, making
//it a complete word.
}
}
The output I'm expecting to get is
3 words found.
9 nodes found.
Instead I'm getting
3 words found.
11 nodes found.
So pretty much a node is being created for ever single letter.
So my assignment requires us to use doubly linked lists to add or multiply numbers together and print them out. I was able to get it to work for whole numbers, but I can't figure out what to change to make it work for decimal numbers as well. Here's what I've got so far. I know it's not the most efficient or cleanest code, but I can try to clarify stuff if it doesn't make sense to you
For example this program will work fine if I do 50382+9281 or 482891*29734,but I need to get it to work for something like 4.9171+49.2917 or 423.135*59
EDIT: Pretend the int values are doubles. I changed it on my actual code, but the result when I do the math is still giving me a whole number so I need to figure out how to insert the decimal at the right place
#include <iostream>
#include <fstream>
#include <string>
#include <stdio.h>
#include <cstdlib>
#include <cstring>
using namespace std;
// A recursive program to add two linked lists
#include <stdlib.h>
#include <assert.h>
#include <math.h>
#include <string.h>
// A linked List Node
struct node
{
int data;
node* next;
node *prev;
};
typedef struct node node;
class LinkedList{
// public member
public:
// constructor
LinkedList(){
int length = 0;
head = NULL; // set head to NULL
node *n = new node;
n->data = -1;
n->prev = NULL;
head = n;
tail = n;
}
// This prepends a new value at the beginning of the list
void addValue(int val){
node *n = new node(); // create new Node
n->data = val; // set value
n->prev = tail; // make the node point to the next node.
// head->next = n;
// head = n;
// tail->next = n; // If the list is empty, this is NULL, so the end of the list --> OK
tail = n; // last but not least, make the head point at the new node.
}
void PrintForward(){
node* temp = head;
while(temp->next != NULL){
cout << temp->data;
temp = temp->next;
}
cout << '\n';
}
void PrintReverse(){
node* temp = tail;
while(temp->prev != NULL){
cout << temp->data;
temp = temp->prev;
}
cout << '\n';
}
void PrintReverse(node* in){
node* temp = in;
if(temp->prev== NULL){
if(temp->data == -1)
cout << temp->data << '\n';
}
else{
cout << temp->data << '\n';
temp = temp->prev;
PrintReverse(temp);
}
}
// returns the first element in the list and deletes the Node.
// caution, no error-checking here!
int popValue(){
node *n = head;
int ret = n->data;
head = head->next;
delete n;
return ret;
}
void swapN(node** a, node**b){
node*t = *a;
*a = *b;
*b = t;
}
node *head;
node *tail;
// Node *n;
};
/* A utility function to insert a node at the beginning of 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;
}
/* A utility function to print linked list */
void printList(struct node *node)
{
while (node != NULL)
{
printf("%d", node->data);
node = node->next;
}
// printf("\n");
}
// A utility function to swap two pointers
void swapPointer( node** a, node** b )
{
node* t = *a;
*a = *b;
*b = t;
}
/* A utility function to get size of linked list */
int getSize(struct node *node)
{
int size = 0;
while (node != NULL)
{
node = node->next;
size++;
}
return size;
}
// Adds two linked lists of same size represented by head1 and head2 and returns
// head of the resultant linked list. Carry is propagated while returning from
// the recursion
node* addSameSize(node* head1, node* head2, int* carry)
{
// Since the function assumes linked lists are of same size,
// check any of the two head pointers
if (head1 == NULL)
return NULL;
int sum;
// Allocate memory for sum node of current two nodes
node* result = (node *)malloc(sizeof(node));
// Recursively add remaining nodes and get the carry
result->next = addSameSize(head1->next, head2->next, carry);
// add digits of current nodes and propagated carry
sum = head1->data + head2->data + *carry;
*carry = sum / 10;
sum = sum % 10;
// Assigne the sum to current node of resultant list
result->data = sum;
return result;
}
// This function is called after the smaller list is added to the bigger
// lists's sublist of same size. Once the right sublist is added, the carry
// must be added toe left side of larger list to get the final result.
void addCarryToRemaining(node* head1, node* cur, int* carry, node** result)
{
int sum;
// If diff. number of nodes are not traversed, add carry
if (head1 != cur)
{
addCarryToRemaining(head1->next, cur, carry, result);
sum = head1->data + *carry;
*carry = sum/10;
sum %= 10;
// add this node to the front of the result
push(result, sum);
}
}
// The main function that adds two linked lists represented by head1 and head2.
// The sum of two lists is stored in a list referred by result
void addList(node* head1, node* head2, node** result)
{
node *cur;
// first list is empty
if (head1 == NULL)
{
*result = head2;
return;
}
// second list is empty
else if (head2 == NULL)
{
*result = head1;
return;
}
int size1 = getSize(head1);
int size2 = getSize(head2) ;
int carry = 0;
// Add same size lists
if (size1 == size2)
*result = addSameSize(head1, head2, &carry);
else
{
int diff = abs(size1 - size2);
// First list should always be larger than second list.
// If not, swap pointers
if (size1 < size2)
swapPointer(&head1, &head2);
// move diff. number of nodes in first list
for (cur = head1; diff--; cur = cur->next);
// get addition of same size lists
*result = addSameSize(cur, head2, &carry);
// get addition of remaining first list and carry
addCarryToRemaining(head1, cur, &carry, result);
}
// if some carry is still there, add a new node to the fron of
// the result list. e.g. 999 and 87
if (carry)
push(result, carry);
}
node* reverse_list(node *m)
{
node *next = NULL;
node *p = m;
node *prev;
while (p != NULL) {
prev = p->prev;
p->prev = next;
next = p;
p = prev;
}
return prev;
}
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////
void Multiply2(node* n1, node* n2);
int digitsPerNode = 2;
node* result;
node* resultp = result;
node* resultp2 = result;
void Multiply(node* n1, node* n2)
{
if (n2->prev != NULL)
{
Multiply(n1, n2->prev);
}
Multiply2(n1, n2);
resultp2 = resultp = resultp->prev;
}
void Multiply2(node* n1, node* n2)
{
if (n1->prev != NULL)
{
Multiply2(n1->prev, n2);
}
if (resultp2 == NULL)
{
resultp2->data = 0;
result = resultp = resultp2;
}
int m = n1->data * n2->data + resultp2->data;
int carryon = (int)(m / pow(10, digitsPerNode));
resultp2->data = m % (int)pow(10, digitsPerNode);
if (carryon > 0)
{
if (resultp2->prev == NULL)
{
resultp2->prev->data = carryon;
}
else
{
resultp2->prev->data += carryon;
}
}
resultp2 = resultp2->prev;
}
/* int* buffer;
int lenBuffer = 0;
void multiplyHelper(int v, node* , int o);
void addToBuffer(int v, int i);
node* multiply(node* num1, node* num2)
{
if (num1 == NULL || num2 == NULL) return NULL;
int length1 = getSize(num1);
int length2 = getSize(num2);
if (length1 > length2) return multiply(num2, num1);
// initialize buffer
lenBuffer = length1 + length2;
buffer = new int[lenBuffer];
memset(buffer, 0, sizeof(int) * lenBuffer);
// multiply
int offset = 0;
node* anode = num1;
while (anode && anode->data!= -1)
{
multiplyHelper(anode->data, num2, offset);
anode = anode->prev;
offset++;
}
// transfer buffer to a linked list
node* h;
int pos = 0;
while (pos < lenBuffer && buffer[pos] == 0) pos++;
if (pos < lenBuffer)
{
node* temp;
temp->data = buffer[pos++];
h = temp;
anode = h;
while (pos < lenBuffer)
{
node* temp;
temp->data = buffer[pos++];
anode->prev = temp;
anode = anode->prev;
}
}
delete buffer;
lenBuffer = 0;
buffer = NULL;
cout << h->data << endl;
return h;
}
// multiply a single digit with a number
// called by multiply()
void multiplyHelper(int value, node* head, int offset)
{
// assert(value >= 0 && value <= 9 && head != NULL);
if (value == 0) return;
node* anode = head;
int pos = 0;
while (anode != NULL)
{
int temp = value * anode->data;
int ones = temp % 10;
if (ones != 0) addToBuffer(ones, offset + pos + 1);
int tens = temp / 10;
if (tens != 0) addToBuffer(tens, offset + pos);
anode = anode->prev;
cout << anode->data;
pos++;
}
}
// add a single digit to the buffer at place of index
// called by multiplyHelper()
void addToBuffer(int value, int index)
{
// assert(value >= 0 && value <= 9);
while (value > 0 && index >= 0)
{
int temp = buffer[index] + value;
buffer[index] = temp % 10;
value = temp / 10;
index--;
}
}*/
// Driver program to test above functions
int main(int argc, char *argv[])
{
char filename[50];
string name= argv[1];
string dig;
name.erase(0,9);//Parse input to only get input file.
ifstream file;
int digits;
for(int i = 0; i < name.length(); i++){
if(name.at(i) == ';'){
// dig = name.substr(0,name.length()-i);
name = name.substr(0,name.length()-i);
}
}
//cout << dig << endl;
//file.open("input.txt");
file.open(name.c_str());
digits = 2;
///////
///////////////////////////////////////////////////////////////////////
int words = 0;
int numbers = 0;
while(!file.eof()) //Goes through whole file until no more entries to input
{
string word;
getline(file,word); //Inputs next element as a string
// word << file;
//cout << word << '\n';
int x = 0;
node *head1 = NULL, *head2 = NULL, *result = NULL;
int counter = 0;
int t1index = 0; //keep tracks of nodes to multiply
int t2index = 0;
char operatorX;
LinkedList tempList1;
LinkedList tempList2;
while(x<word.length()) //Loops through each string input
{
//if(x<word.length()&&isalpha(word.at(x))) //Checks that x is in bounds and that char at position x is a letter
if(x<word.length()&&isdigit(word.at(x))) //Checks that x is in bounds and that char at position x is a number/digit
{
int start = x;
while(x<word.length()&&isdigit(word.at(x))) //Loops past the number portion
{
x++;
}
string temp = word.substr(start, x).c_str();
// cout << temp << '\n';
for(int i = 0; i < temp.length();i++){
tempList1.addValue(atoi(temp.substr(i, 1).c_str()));
// push(&head1, atoi(temp.substr(i, 1).c_str()));
counter++;
t1index++;
}
//search for the operator
while(x<word.length()){
if(x<word.length()&& (!isspace(word.at(x)) && !isdigit(word.at(x))))
{
while(x<word.length()&&(!isspace(word.at(x)) && !isdigit(word.at(x)))) //Loops past the letter portion
{
// cout << (word.at(x))<< '\n';
operatorX = word.at(x);
x++;
}
//search second value
while(x<word.length()){ //second value find
//start
if(x<word.length()&&isdigit(word.at(x))) //Checks that x is in bounds and that char at position x is a number/digit
{
int start = x;
while(x<word.length()&&isdigit(word.at(x))) //Loops past the number portion
{
x++;
}
string temp = word.substr(start, x).c_str();
for(int i = 0; i < temp.length();i++){
tempList2.addValue(atoi(temp.substr(i, 1).c_str()));
// push(&head2, atoi(temp.substr(i, 1).c_str()));
// cout << atoi(temp.substr(i, 1).c_str());
counter++;
}
//////START READING NUMBERS BACKWARDS
LinkedList finalList;
node* tempA = tempList1.tail;
node* tempB = tempList2.tail;
// multiply(tempA, tempB);
//ADDITION
while(tempA != NULL){
if(tempA->data != -1){
push(&head1,tempA->data);
// cout << tempA->data;
}
tempA = tempA->prev;
}
while(tempB != NULL){
if(tempB->data != -1){
push(&head2, tempB->data);
// cout << tempB->data;
}
tempB = tempB->prev;
}
// multiply(head1, head2);
// result = multiply(head1, head2);
// tempList1.PrintReverse();
addList(head1, head2, &result);
printList(head1);
cout << operatorX;
printList(head2);
cout << "=";
printList(result);
cout << endl;
}
else{
x++;
}
//end
}
}
else{
x++;
}
}
}
else //If char at position x is neither number or letter skip over it
{
x++;
}
}
}
}
Since you're working in C++, use a template/overloaded operators. Cast your ints to a floating point type as necessary. See e.g.:
C++ Template problem adding two data types
Hey I have to find the most eficient way to print a number by giving the postion. The input is like this:
8 (N-> N Numbers)
INS 100 (Add 100 to the tree)
INS 200 (Add 200 to the tree)
INS 300 (Add 300 to the tree)
REM 200 (Remove the number 200 from the tree)
PER 1 (Have to output the biggest number in the tree-> Shoud print 300)
INS 1000 (Add 1000 to the tree)
PER 1 ((Have to output the biggest number in the tree-> Shoud print 1000))
PER 2 (I have to output the second biggest number so: 300)
I have a way to print like this, but is very slow and I have to maintain a O(N * log(N)).
Here is my full code
#include<stdio.h>
#include<stdlib.h>
#include<iostream>
using namespace std;
// An AVL tree node
struct node
{
int key;
struct node *left;
struct node *right;
int height;
};
// A utility function to get maximum of two integers
int max(int a, int b);
// A utility function to get height of the tree
int height(struct node *N)
{
if (N == NULL)
return 0;
return N->height;
}
// A utility function to get maximum of two integers
int max(int a, int b)
{
return (a > b)? a : b;
}
/* Helper function that allocates a new node with the given key and
NULL left and right pointers. */
struct node* newNode(int key)
{
struct node* node = (struct node*)
malloc(sizeof(struct node));
node->key = key;
node->left = NULL;
node->right = NULL;
node->height = 1; // new node is initially added at leaf
return(node);
}
// A utility function to right rotate subtree rooted with y
// See the diagram given above.
struct node *rightRotate(struct node *y)
{
struct node *x = y->left;
struct node *T2 = x->right;
// Perform rotation
x->right = y;
y->left = T2;
// Update heights
y->height = max(height(y->left), height(y->right))+1;
x->height = max(height(x->left), height(x->right))+1;
// Return new root
return x;
}
// A utility function to left rotate subtree rooted with x
// See the diagram given above.
struct node *leftRotate(struct node *x)
{
struct node *y = x->right;
struct node *T2 = y->left;
// Perform rotation
y->left = x;
x->right = T2;
// Update heights
x->height = max(height(x->left), height(x->right))+1;
y->height = max(height(y->left), height(y->right))+1;
// Return new root
return y;
}
// Get Balance factor of node N
int getBalance(struct node *N)
{
if (N == NULL)
return 0;
return height(N->left) - height(N->right);
}
struct node* insert(struct node* node, int key)
{
/* 1. Perform the normal BST rotation */
if (node == NULL)
return(newNode(key));
if (key < node->key)
node->left = insert(node->left, key);
else
node->right = insert(node->right, key);
/* 2. Update height of this ancestor node */
node->height = max(height(node->left), height(node->right)) + 1;
/* 3. Get the balance factor of this ancestor node to check whether
this node became unbalanced */
int balance = getBalance(node);
// If this node becomes unbalanced, then there are 4 cases
// Left Left Case
if (balance > 1 && key < node->left->key)
return rightRotate(node);
// Right Right Case
if (balance < -1 && key > node->right->key)
return leftRotate(node);
// Left Right Case
if (balance > 1 && key > node->left->key)
{
node->left = leftRotate(node->left);
return rightRotate(node);
}
// Right Left Case
if (balance < -1 && key < node->right->key)
{
node->right = rightRotate(node->right);
return leftRotate(node);
}
/* return the (unchanged) node pointer */
return node;
}
/* Given a non-empty binary search tree, return the node with minimum
key value found in that tree. Note that the entire tree does not
need to be searched. */
struct node * minValueNode(struct node* node)
{
struct node* current = node;
/* loop down to find the leftmost leaf */
while (current->left != NULL)
current = current->left;
return current;
}
struct node* apagaNode(struct node* root, int key)
{
// STEP 1: PERFORM STANDARD BST DELETE
if (root == NULL)
return root;
// If the key to be deleted is smaller than the root's key,
// then it lies in left subtree
if ( key < root->key )
root->left = apagaNode(root->left, key);
// If the key to be deleted is greater than the root's key,
// then it lies in right subtree
else if( key > root->key )
root->right = apagaNode(root->right, key);
// if key is same as root's key, then This is the node
// to be deleted
else
{
// node with only one child or no child
if( (root->left == NULL) || (root->right == NULL) )
{
struct node *temp = root->left ? root->left : root->right;
// No child case
if(temp == NULL)
{
temp = root;
root = NULL;
}
else // One child case
*root = *temp; // Copy the contents of the non-empty child
free(temp);
}
else
{
// node with two children: Get the inorder successor (smallest
// in the right subtree)
struct node* temp = minValueNode(root->right);
// Copy the inorder successor's data to this node
root->key = temp->key;
// Delete the inorder successor
root->right = apagaNode(root->right, temp->key);
}
}
// If the tree had only one node then return
if (root == NULL)
return root;
// STEP 2: UPDATE HEIGHT OF THE CURRENT NODE
root->height = max(height(root->left), height(root->right)) + 1;
// STEP 3: GET THE BALANCE FACTOR OF THIS NODE (to check whether
// this node became unbalanced)
int balance = getBalance(root);
// If this node becomes unbalanced, then there are 4 cases
// Left Left Case
if (balance > 1 && getBalance(root->left) >= 0)
return rightRotate(root);
// Left Right Case
if (balance > 1 && getBalance(root->left) < 0)
{
root->left = leftRotate(root->left);
return rightRotate(root);
}
// Right Right Case
if (balance < -1 && getBalance(root->right) <= 0)
return leftRotate(root);
// Right Left Case
if (balance < -1 && getBalance(root->right) > 0)
{
root->right = rightRotate(root->right);
return leftRotate(root);
}
return root;
}
int imprime(struct node *root,int targetPos,int curPos)
{
if(root != NULL)
{
int newPos = imprime(root->left, targetPos, curPos);
newPos++;
if (newPos == targetPos)
{
printf("%d\n", root->key);
}
return imprime(root->right, targetPos, newPos);
}
else
{
return curPos;
}
}
int main()
{
struct node *root = NULL;
int total=0;
int n,b;
string a;
cin >> n;
for (int i=0; i<n; i++)
{
cin >> a >> b;
if(a=="INS")
{root = insert(root, b);total=total+1;}
else
if(a=="REM")
{root = apagaNode(root, b);total=total-1;}
else
imprime(root, total-b+1, 0);
}
return 0;
}
The way I found to print the values:
int imprime(struct node *root,int targetPos,int curPos)
{
if(root != NULL)
{
int newPos = imprime(root->left, targetPos, curPos);
newPos++;
if (newPos == targetPos)
{
printf("%d\n", root->key);
}
return imprime(root->right, targetPos, newPos);
}
else
{
return curPos;
}
}
Problem is that this function is very slow, and I can't use it. How is the best way to print by a given postion like this? (A heard about, counting n_nodes, and during the rotations I have to incremennt, decrement, i realy did not understand. Help me please! Give some tips, and advices) (PS: I'm not an expert with this kind of algorithms)
The advice you heard is correct: you should add a node counter to your node structure:
struct node
{
int key;
struct node *left;
struct node *right;
int height;
int n_nodes;
};
It should hold the number of nodes in the tree. Assuming it's correct, you can improve the algorithm for finding a node with a target position: it will know exactly in which branch of the tree to look (left or right), which will make the search faster (current imprime implementation is O(n)).
So, how to make it so the n_nodes field holds the right value? Fortunately, you already have an example: height. Look where your existing code changes it; these are roughly the places where you have to update n_nodes, too. Most of them are trivial (just add 1 to it); the more interesting ones are the rotation functions:
struct node *rightRotate(struct node *y)
{
struct node *x = y->left;
struct node *T2 = x->right;
// Perform rotation
x->right = y;
y->left = T2;
// Update heights
y->height = max(height(y->left), height(y->right))+1;
x->height = max(height(x->left), height(x->right))+1;
// Update numbers of nodes
x->n_nodes = ...;
y->n_nodes = ...;
T2->n_nodes = ...;
// Return new root
return x;
}
So it transforms the tree like this:
y x
/ \ / \
x D A y
/ \ ==> / \
A T2 T2 D
/ \ / \
B C B C
Here A, B, C and D are trees whose sizes your program knows; let's denote their sizes as a, b, c and d. So the transformation changes these sizes like this:
size of x: from a+b+c+2 to a+b+c+d+3
size of y: from a+b+c+d+3 to b+c+d+2
size of T2: unchanged
So just transform this to code.
so heres my header and cpp file.
template<typename T> struct TreeNode
{
TreeNode(const T& value, TreeNode<T>* left = NULL, TreeNode<T>* right = NULL)
{
Value = value;
Left = left;
Right = right;
}
T Value;
TreeNode<T>* Left;
TreeNode<T>* Right;
bool IsLeaf() const
{
return Left == NULL && Right == NULL;
}
};
and now my cpp file
#include "TreeNode.h"
#include <iostream>
#include <string>
using namespace std;
float ValueOf(TreeNode<char>* root);
float ValueOf(TreeNode<char>* root)
{
if (root->IsLeaf())
return root->Value - '0';
float expressionValueL = ValueOf(root->Left);
float expressionValueR = ValueOf(root->Right);
if (root->Value == '+')
return expressionValueL+expressionValueR;
else if (root->Value == '-')
return expressionValueL-expressionValueR;
else if (root->Value == '*')
return expressionValueL*expressionValueR;
else if (root->Value == '/')
return expressionValueL/expressionValueR;
}
void main ()
{
TreeNode<char>* treeRoot = nullptr;
TreeNode<char>* currentNode = treeRoot;
string expr;
cout<<"please input expression to be tested:";
getline (cin, expr);
cout<<endl;
int size = expr.size();
for (int i=0; i<size; i++)
{
char test = expr[i];
if ((test=='1')||(test=='0')||(test=='2')||(test=='3')||(test=='4')||(test=='5')||(test=='6')||(test=='7')||(test=='8')||(test=='9'))
{
TreeNode<char> newLeaf = (expr[i]);
if (currentNode == nullptr)
{
treeRoot=&newLeaf;
currentNode = &newLeaf;
}
else
currentNode->Right = &newLeaf;
}
else if ((expr[i]=='+')||(expr[i]=='-'))
{
TreeNode<char> newRoot = test;
newRoot.Left = treeRoot;
treeRoot = &newRoot;
currentNode = &newRoot;
}
else if (((expr[i]=='*')||(expr[i]=='/'))&&(currentNode->Right==nullptr))
{
TreeNode<char> newRoot = test;
newRoot.Left = treeRoot;
treeRoot = &newRoot;
currentNode = &newRoot;
}
else if (((expr[i]=='*')||(expr[i]=='/'))&&(currentNode->Right!=nullptr))
{
TreeNode<char> newChild = test;
newChild.Left = currentNode->Right;
currentNode->Right = &newChild;
currentNode = &newChild;
}
}
cout<<ValueOf(treeRoot)<<endl;
system("pause");
}
the problem is that every time i run it, and i input something like 3*4-2, all of the digits in the tree gets overwritten to what the last digit inserted was, so its interpreted as 2*2-2 and gives me 2 as an answer, instead of 10 can anyone tell me what my problem is? thanks =).
btw this program assumes wellformed expressions and single digit numbers.
TreeNode<char> newLeaf = (expr[i]); creates an object on stack - it's invalidated when you leave the enclosing scope. You should not store pointers to such objects.
Use TreeNode<char> * newLeaf = new TreeNode<char>(expr[i]); - and corresponding for any other node that you assign to ->Right and ->Left - aka need to keep alive beyond the scope where you create them.
As mentioned by Erik,
TreeNode<char> newLeaf = (expr[i]);
is a local stack variable; instead of do following:
TreeNode<char> *newLeaf = new TreeNode<char>(expr[i]);
And then assign to proper leg. Also, below condition,
if ((test=='1')||(test=='0')||(test=='2')||(test=='3')||(test=='4')||(test=='5')||(test=='6')||(test=='7')||(test=='8')||(test=='9'))
Can be squeezed to,
if(test >= '0' && test <= '9')
In the same way, you can also co-relate the last two else if() statements for better code