I wrote a code that suposed to return the depth of a binary tree from the root to the node who called the function. using recursive way but I faced a problem about how to count the number of times that the function gets called so I whould know how much convexity I passed. Someone know how can I do that?
int BSNode::getDepth(const BSNode& root) const
{
if (this != nullptr)
{
if (root.getData() > this->_data)
{
this->getDepth(*root.getRight());
}
else if (root.getData() < this->_data)
{
this->getDepth(*root.getLeft());
}
else if (root.getData() == this->_data)
{
// return the number that the function counted
}
}
else
{
return 0;
}
}
You should at least return something in every case. And when you arrive at the intended node (having the data you are looking for), then return 0. In all other cases, return what you get from recursion plus 1. If the value is not found then indeed -1 should be returned. And if this -1 is coming back from recursion, it should be returned like that also to the caller (without adding 1).
Here is the code adapted:
int BSNode::getDepth(const BSNode& root) const
{
int temp;
if (this != nullptr)
{
if (root.getData() > this->_data)
{
temp = this->getDepth(*root.getRight());
return temp == -1 ? -1 : temp + 1;
}
else if (root.getData() < this->_data)
{
temp = this->getDepth(*root.getLeft());
return temp == -1 ? -1 : temp + 1;
}
else if (root.getData() == this->_data)
{
return 0;
}
}
else
{
return -1;
}
}
Related
I am trying traverse over tree and check for specific (rel_name) and return his height, but my function traverse over "mother" branch and checks only fathers branch.
The result is that my program return exception() and core dumping.
how do I fix my function to not core dump and check mothers branch too?
string treeHeight(Person* root, string rel_name, int height){
height++;
if(root == nullptr) {
throw exception();
}
else if(root->name == rel_name) return to_string(height);
return treeHeight(root->father, rel_name, height);
return treeHeight(root->mother, rel_name, height);
}
Someone has already pointed out, but if your code reaches a leaf and goes a little bit further, according to your base case it would throw an exception and exit the program.
I revised your code a little bit:
int treeHeight(Person* root, const string& rel_name){
if(root == nullptr) {
return -1;
}
else if(root->name == rel_name) return 0;
int leftHeight = treeHeight(root->father, rel_name);
int rightHeight = treeHeight(root->mother, rel_name);
if (leftHeight == -1 && rightHeight == -1) {
return -1;
} else {
return (leftHeight > rightHeight ? leftHeight : rightHeight) + 1;
}
}
Hope it helps.
In your code , you are not allowing the code to go the mother node, its directly from the father node call and thus its not getting executed ahead.
Try this below :
string treeHeight(Person* root, string rel_name, int height){
height++;
if(root == nullptr) {
throw exception();
}
else if(root->name == rel_name) return to_string(height);
int person_height = treeHeight(root->father, rel_name, height);
if(person_height!=0) return height; ---------> You need to apply this check
return treeHeight(root->mother, rel_name, height);
}
void head_insert(DomesticPtr& head, string firstNameD, string lastNameD, string province, float cgpaD, int researchScoreD, int idD)
{
DomesticPtr temp_ptr;
DomesticPtr temp2;
temp_ptr= new DomesticStudent(head, firstNameD, lastNameD, province,cgpaD,researchScoreD,idD);
temp2 = head->getLink();
temp2==temp_ptr;
head=temp_ptr;
if (head->getLink() == NULL)
return;
else
{
bubblesort(head);
}
}
void bubblesort(DomesticStudent* head)
{
int rsd;
int cgpad;
int p;
DomesticPtr tempc, tempd, tempe;
tempd=head;
tempe= head->getLink();
{
while(tempd != NULL)
{
rsd=compareResearchScore(tempd, tempe);
if (rsd==1)
{
tempc=head;
head->next=head;
head=tempc;
}// if
else if (rsd==0)
{
cgpad= compareCGPA(tempe,tempd);
if (cgpad==1)
{
tempc=head;
head->next=head;
head=tempc;
}// if (cgpad[k]>cgpad[k+1])
else if(cgpad==0)
{
p=compareProvince(tempd,tempe);
if(p==1)
{
tempc=head;
head->next=head;
head=tempc;
}// if (p[k]>p[k+1])
}//
}// else if cgpad[k]
}// else if rsd[k]
// }
// }
tempd = tempe;
}
int compareResearchScore(DomesticPtr RSA, DomesticPtr RSB)
{
if (RSB == NULL || RSA==NULL )
{
return 0;
}
if (RSA->researchScoreD==RSB->researchScoreD) //compares if is the same for domesetic students returns value for bubble sort
{
return 0;
}
if (RSA->researchScoreD > RSB->researchScoreD)
{
return 1;
}
if (RSA->researchScoreD< RSB->researchScoreD)
{
return -1;
}
}
I'm trying to to have my linked list sorted every time a new node is inserted. It compiles but every time I try to run the program it is stuck on the point that I am trying to print my list. I have a destructor but no copy constructor or assignment operator.
The head_insert calls the sort function and the sort function calls the compare function to receive an integer output so that it can make a swap. I want to compare research, the cgpa, and then province. Any input would be much appreciated, this is for a project so I wouldn't like any blocks of code but if you could point me in the right direction or multiple directions.
I am working on an assignment in which we must create a 20-questions type game from a binary search tree. We read the tree in from a text file that is formatted like this:
Does it walk on 4 legs?
Does it fly?
*centipede?
Is it an insect?
*bird?
*butterfly?
Does it purr?
Does it howl?
*mouse?
*dog?
*cat?
Later, I am going to allow the user to add to this list. At the moment, however, I am unable to accurately read the list into a binary search tree. I have set it up so that (I think) it will use recursion and return to the previous "current" node pointer when it ends a loop of the function. Currently, however, the current node pointer remains the same.
The below function is passed a vector of the strings from the text file.
string line;
string guess;
bool start = true;
void buildTree(vector<string> gameData, Node* current, int &counter)
{
//fill node with question or answer
//recursive:
// add to the left until we encounter an asterisk
// add to the right
line = gameData[counter];
//if a question
if (line[0] != '*')
{
if (current->getData().empty())
{
current->setData(line);
cout << current->getData() << endl;
}
if (!start)
{
//if noChild is empty AND current isn't a guess, go to noChild
if ((current->getNo()->getData().empty())
&& (current->isGuess() == false))
{
current = current->getNo();
}
//otherwise, go to yes
else {
current = current->getYes();
}
}
while (counter < gameData.size())
{
if (!start) { counter++; }
start = false;
buildTree(gameData, current, counter);
}
}
//if a guess
else
{
//if data is full, go to no
if (current->getData().empty() == false)
{
current = current->getNo();
}
//otherwise, go to yes
else
{
//current = current->getYes();
for (int i = 1; i < line.size(); i++)
{
guess.push_back(line[i]);
}
current->setData(guess);
guess.clear();
cout << current->getData() << endl;
counter++;
current->setGuess(true);
}
}
}
I cannot figure out why this error is happening: error: "control may reach end of non-void function" even when "else" statement is present at the end.
Here is the code:
bnode* binsert(bnode *h,int k){
bnode *temp=new bnode;
if(h==NULL)
{
temp->num=k;
temp->L=NULL;
temp->R=NULL;
h=temp;
return h;
}
else if(h->L==NULL && k<h->num)
{
temp->num=k;
temp->L=NULL;
temp->R=NULL;
h->L=temp;
return h;
}
else if(h->R==NULL && k>h->num)
{
temp->num=k;
temp->L=NULL;
temp->R=NULL;
h->R=temp;
return h;
}
else if(h->L!=NULL && k<h->num)
{
h->L=binsert(h->L,k);
}
else
{
h->R=binsert(h->R,k);
}
}
You need to return the results of recursive calls, it's not done automatically.
You can also simplify your code a bit by adding a constructor:
bnode::bnode(int v)
: num(v),
L(nullptr),
R(nullptr)
{
}
and since you're already handling the case of a null parameter, you don't need special cases for null children:
bnode* binsert(bnode *h,int k)
{
if(h == nullptr)
{
h = new bnode(k);
}
else if(k < h->num)
{
h->L = binsert(h->L, k);
}
else if(k > h->num)
{
h->R = binsert(h->R, k);
}
return h;
}
because this last 2 conditions:
else if(h->L!=NULL && k<h->num)
{
h->L=binsert(h->L,k);
}
else
{
h->R=binsert(h->R,k);
}
may occur and no return is given...
you need to be sure the function returns a value no matter what the condition evaluates....
else if(h->L!=NULL && k<h->num)
{
h->L=binsert(h->L,k);
}
else
{
h->R=binsert(h->R,k);
}
In the else if and else cases for your code, if you reach here, you do not return a value, and the behavior is undefined if you try to use this value.
You probably want to add a return h; in the two branches.
I have this challenging exercise I got from a book about c++, and i'm not sure how to tackle this problem.
I must define a function called treeNodeCount() which returns the number of nodes in a binary tree (easy enough), and I also have to define an overloaded function that takes an int(0,1, or 2) which represents the number of children, and the function should return the nodes that have that specific number of children.
treeNodeCount should both use a function called nodeCount(elemType root) to do the recursion necessary to count the nodes(so basically all the work).
challenge number one says that I can add a second parameter to nodeCount which takes the number of children for the nodes that we want to count.
Challenge number two says that we cannot use a second parameter (this is the tough part)
I was able to do challenge one and here is what I came up with:
template <class elemType>
int binaryTreeType<elemType>::nodeCount(nodeType<elemType> *p, int a ) const
{
if (p == NULL){
return 0;
}
else if (a == 0 && p->lLink == NULL && p->rLink == NULL){
return 1 + nodeCount(p->lLink, a) + nodeCount(p->rLink, a);
}
else if (a == 1 && (p->lLink != NULL ^ p->rLink != NULL)){
return 1 + nodeCount(p->lLink, a) + nodeCount(p->rLink, a);
}
else if (a == 2 && p->lLink != NULL && p->rLink != NULL){
return 1 + nodeCount(p->lLink, a) + nodeCount(p->rLink, a);
}
else if (a == -1){
return nodeCount(p->lLink, a) + nodeCount(p->rLink, a) + 1;
}
return nodeCount(p->lLink, a) + nodeCount(p->rLink, a);
}
template <class elemType>
int binaryTreeType<elemType>::treeNodeCount(int a) const{
return nodeCount(root, a);
}
This seems to work fine but i am convinced that there has to be a better way.
I was not able to do challenge 2 though, and i have no idea what to do (is it even possible)
You can scrunch down your logic and make it a bit more straightforward by implementing a function to return the number of children given a node.
template <class elemType>
int nodeSize(nodeType<elemType>* node) const
{
int count = 0;
if (node->lLink)
++count;
if (node->rLink)
++count;
return count;
}
template <class elemType>
int binaryTreeType<elemType>::nodeCount(nodeType<elemType>* node, int count) const
{
if (node)
{
if (nodeSize(node) == count || count == -1)
return nodeCount(node->lLink, count) + nodeCount(node->rLink, count) + 1;
return nodeCount(node->lLink, count) + nodeCount(node->rLink, count);
}
return 0;
}
For the second challenge, you need a stack to avoid recursion.
template <class elemType>
int binaryTreeType<elemType>::treeNodeCount(int count) const
{
stack<nodeType<elemType>*> node_stack;
node_stack.push(root);
int num_matches = 0;
while (!stack.empty())
{
nodeType<elemType>* node = node_stack.top();
node_stack.pop();
if (node)
{
if (nodeSize(node) == count || count == -1)
++num_matches;
node_stack.push(node->lLink);
node_stack.push(node->rLink);
}
}
return num_matches;
}
Edit: fixed a goof in the above recursive version. Thanks to David Rodriguez for pointing it out.