Hi this is a code from my SearchTree class.
Node* is a structure whith m_info type int, and m_left(smaller nodes by info) and m_right(bigger nodes by info)
void SearchTree::insert(const int &x) {
Node* tempo = m_root;
while (tempo != nullptr) {
if (tempo->m_info >= x) {
tempo = tempo->m_left;
} else {
tempo = tempo->m_right;
}
}
tempo = new Node(x);
}
I am trying to insert a new node to the tree.
But looks like I am missing something in memory management.
There tempo is a pointer to a new node, however it is not being related to m_root.
I am confused here. I really love the power of c++ but it bends my logic.
What am I missing here?
You keep advancing tempo until it is equal to nullptr. At this point you have left the tree and all you have in hand is a pointer into nothingness. Note that in particular the program has no way of determining which node you last visited that led to tempo becoming null.
What you need to do instead is stop one step earlier: While tempo is still pointing to a node, but the next step would make it point to null. Now you still have a valid node of the tree in your hand and can attach the newly allocated node to it.
You can't save the pointer in tempo only. Tempo is a copy of your current position in the tree. You have to assign it to actual variable.
My solution to this problem would be to check if child is nullptr before you iterate
void SearchTree::insert(const int &x) {
if (!m_root) {
m_root = new Node(x);
return;
}
Node* tempo = m_root;
while (true) {
if (tempo->m_info >= x) {
if (!tempo->m_left) {
tempo->m_left = new Node(x);
return;
}
tempo = tempo->m_left;
} else {
if (!tempo->m_right) {
tempo->m_right = new Node(x);
return;
}
tempo = tempo->m_right;
}
}
}
Also you should use smart pointers instead of raw pointers.
An alternative solution is a pointer to pointer. I didn't test it but you can try
void SearchTree::insert(const int &x) {
Node** tempo = &m_root;
while (*tempo) {
if ((*tempo)->m_info >= x) {
tempo = &(*tempo)->m_left;
} else {
tempo = &(*tempo)->m_right;
}
}
*tempo = new Node(x);
}
In this image you can see. If you use Node* tempo = m_root then tempo contains a copy of the value in m_root. If you change tempo then m_root stays unchanged.
If you use Node** tempo = &m_root then tempo is pointer to m_root. You can change m_root through tempo.
Related
I have been trying to solve the selection sort in single linked list without using swap nodes. Using a temp list to store nodes and assign the current list with a new one
//my addlastnode function
void AddLastNODE(LIST &mylist, NODE *p)
{
//Check the list is empty or not
if(isEmpty(mylist))
mylist.pHead = mylist.pTail = p;
else
mylist.pTail->pNext = p;
mylist.pTail = p;
}
void selectionSort(LIST &mylist)
{
//Initialize a temp list to store nodes
LIST mylisttemp;
IntList(mylisttemp);
//Create node
NODE *p;
NODE *i;
//Create min node
NODE *min;
//Check if list is empty or has one node
if(mylist.pHead == mylist.pTail)
return;
//Traverse the list till the last node
for(p=mylist.pHead; p->pNext!=NULL && p!=NULL; p = p->pNext)
{
min=p;
for(i=p->pNext; i!=NULL;i=i->pNext)
{
////Find the smallest data in list
if(i->data < min->data)
min=i;
}
////Add the smallest to a new list
AddLastNODE(mylisttemp, min);
}
//Fill the current list to the new list
if(!isEmpty(mylisttemp))
mylist = mylisttemp;
}
Your code does not reduce the list you are selecting nodes from: the selected node should be removed from it. To make that happen, you need a reference to the node before the selected node, so that you can rewire the list to exclude that selected node.
There is also a small issue in your AddLastNODE function: it does not force the tail node to have a null as pNext pointer. This may be a cause of errors when the function is called with a node that still has a non-null pNext pointer. Secondly, the indentation is off around the else block. It does not lead to a bug in this case, but still it is better to avoid the confusion:
void AddLastNODE(LIST &mylist, NODE *p)
{
if(isEmpty(mylist))
mylist.pHead = p;
else
mylist.pTail->pNext = p;
mylist.pTail = p; // indentation!
p->pNext = nullptr; // <--- better safe than sorry!
}
Then to the main algorithm. It is quite tedious to work with a previous node reference when looking for the node with the minimum value. It helps a bit when you temporarily make the input list cyclic:
void selectionSort(LIST &mylist) {
if (mylist.pHead == mylist.pTail) return;
// Make list temporarily cyclic
mylist.pTail->pNext = mylist.pHead;
LIST mytemplist;
IntList(mytemplist);
while (mylist.pHead != mylist.pTail) {
// Select node:
NODE * beforemin = mylist.pTail;
for (NODE * prev = mylist.pHead; prev != mylist.pTail; prev = prev->pNext) {
if (prev->pNext->data < beforemin->pNext->data) {
beforemin = prev;
}
}
NODE * min = beforemin->pNext;
// Extract selected node:
if (min == mylist.pTail) mylist.pTail = beforemin;
if (min == mylist.pHead) mylist.pHead = min->pNext;
beforemin->pNext = min->pNext;
// And insert it:
AddLastNODE(mytemplist, min);
}
// Move last remaining node
AddLastNODE(mytemplist, mylist.pHead);
// Copy back
mylist = mytemplist;
}
As a side note: You might even want to always keep your list cyclic. This will mean some changes in other functions you may have, as there will be no pNext pointers that are null then.
This what I need to do:
append_data adds a node onto the end of the list pointed to by top. the resulting list is one element longer, and the newly appended node has the given data value. consider using the 'append' function to help.
void AppendData(int data);
append is the same as append_data, except we're adding a node, rather than a value.
void Append(shared_ptr new_node);
I was able to AppendData using this code, but in an ideal world I could pass Append thru AppendData and get the same result. I'm struggling with Append right now even though the answer is in the AppendData code
'''
void LinkedList::AppendData(int data){
shared_ptr<node> temp(new node);
temp->data = data;
temp->next = shared_ptr<node>(NULL);
shared_ptr<node> end_node(top_ptr_);
if(end_node == NULL) {
end_node=temp;
} else {
while(end_node->next!=NULL) {
end_node=end_node->next;
}
end_node->next=temp;
}
}
void LinkedList::Append(shared_ptr<node> new_node){}
'''
in an ideal world I could pass Append thru AppendData and get the same result.
Actually, Append has less information than AppendData, in particular, the data itself. It would make more sense to refactor it in a way that AppendData creates a node, and then calls Append to append that node to the list:
void LinkedList::AppendData(int data){
// construct node
shared_ptr<node> temp(new node);
temp->data = data;
temp->next = shared_ptr<node>(NULL);
// append it
Append(temp);
}
void LinkedList::Append(shared_ptr<node> new_node){
shared_ptr<node> end_node(top_ptr_);
if(end_node == NULL) {
end_node = new_node; // new_node instead of temp
} else {
while(end_node->next!=NULL) {
end_node=end_node->next;
}
end_node->next = new_node; // ditto
}
}
Also, this line end_node = new_node; is incorrect. If the list is empty, you need to update top_ptr_ like this:
top_ptr_ = new_node;
The backPtr_ of last node is pointing to itself at the time of generating trajectory.
This is mostly happening due to the backptr__ is pointing at the back of openNodes and while exploring a node it is last element due to pop_heap.
Any work around for this?
I have already tried to keep the openNodes as vector of Node objects rather than of pointers.
init_node.calculateF();
vector<Node*> openNodes;
openNodes.push_back(&init_node);
vector<vector<coord_t>> traj;
while(!openNodes.empty()){
pop_heap(openNodes.begin(), openNodes.end(), NodePtrCompareGtr());
Node* newNodePtr = openNodes.back();
Node newNode = *newNodePtr;
openNodes.pop_back();
traj = newNode.exploreNodes(openNodes, openSet, closedNodes, &newNode);
}
class Node
{
Node *backPtr_ = NULL;
public:
void setBackPtr(Node* nodeptr){
backPtr_ = nodeptr;
}
vector<vector<coord_t>> exploreNodes(vector<Node*> &openNodes){
newCoordList = generateCoords();
vector<Node*> newNodeList;
if(goal_reached(this)){
return generate_traj(this);
}
for(auto const& newCoords : newCoordsList){
Node *newNode = new Node(...);
newNodeList.push_back(newNode);
}
for(auto & i : newNodeList){
...
i->setBackPtr(this);
openNodes.push_back(i);
make_heap(openNodes.begin(), openNodes.end(), NodePtrCompareGtr());
...
}
}
I want to store the backPtr_ correctly to store the correct previous node.
Edited*: I'm working on the delete function for a binary search tree. I'm just working on the first case right now. I think this is correct, but I'm wondering if it can be done recursively, or more efficiently. Any help is appreciated. Assume BSTSearch searches for a node, isLeaf returns true if the node is a leaf, and each node has a pointer that allows them access to their parent.
void
BinarySearchTree::BSTDelete(int x, BSTNode *node){
BSTNode *deleteNode;
deleteNode = BSTSearch(x,node);
if(isLeaf(deleteNode)){
if(deleteNode->sortkey > (deleteNode->parent)->sortkey){
delete (deleteNode->parent)->right;
(deleteNode->parent)->right = NULL;
}
else{
delete (deleteNode->parent)->left;
(deleteNode->parent)->left = NULL;
}
}
You don't need a pointer to the parent. Here is a recursive version that should work: (pass by reference (&), in case you don't know, allows you to modify the variable, similar to pass by pointer; BSTNode *& is a pointer passed by reference, so we can modify the value of node->left/right (pointers) (not just what they're pointing to))
void BinarySearchTree::BSTDelete(int x, BSTNode *&node)
{
if (node == NULL)
return;
if (x == node->sortKey)
{
if (isLeaf(node))
{
delete node;
node = NULL;
}
else
{
// other stuff goes here
}
return;
}
else if (x < node->sortKey)
BSTDelete(x, node->left);
else
BSTDelete(x, node->right);
}
This is an interview question that I found interesting.
Write a method that takes a pointer to a Node structure as a parameter and returns a complete copy of the passed-in data structure.
The Node structure contains two pointers to other Node structures.
For example, the method signature could look like so:
Node* Copy(Node* root);
Note - Do not make any assumptions about the data structure – it could be a tree, linked list, graph, etc.
How can this be done for any data structure ?
In the generic graph case, you need a mapping from nodes in the original graph to nodes in the new graph, so that when a cycle is encountered, the proper link gets created. If you happen to have extra temporary space in each node, large enough to hold a pointer, then you can store the mapping directly in the nodes; otherwise, you'll need to use an external map, such as an associative array or hash table.
Then it's just a matter of traversing the graph, copying nodes, and looking up the corresponding edges. Something like this:
struct Node
{
Node(int _data) : data(_data) { memset(links, 0, sizeof(links)); }
int data;
Node *links[2];
}
Node *Copy(Node *root)
{
typedef std::map<Node*, Node*> NodeMap;
NodeMap nodeMap;
std::deque<Node*> nodesToVisit;
// Set up initial new root and mapping for the root
Node *newRoot = new Node(root->data);
nodeMap[root] = newRoot;
// Breadth-first search the graph
nodesToVisit.push_back(root);
while(!nodesToVisit.empty())
{
Node *cur = nodesToVisit.front();
nodesToVisit.pop_front();
Node *newCur = nodeMap[cur];
for(int i = 0; i < 2; i++)
{
Node *link = cur->links[i];
if(link)
{
// If we've already created the corresponding node for this
// link, use that. Otherwise, create it and add it to the map.
NodeMap::iterator mappedLink = nodeMap.find(link);
if(mappedLink != nodeMap.end())
{
newCur->links[i] = mappedLink->second;
}
else
{
Node *newLink = new Node(link->data);
nodeMap[link] = newLink;
newCur->links[i] = newLink;
nodesToVisit.push_back(link);
}
}
}
}
return newRoot;
}
The problem as stated is impossible. You have to assume that the entire data structure is stored entirely within the content of nodes that are accessible from that initial one. But that is not an assumption you are allowed to make. Even your standard basic double linked list might not fit that description.
class Copier {
std::map <Node*, Node*> copies;
Node* Copy(Node* n) {
if (!n) return 0;
Node*& copy = copies[n];
if (!copy) {
copy = new Node();
copy.node1 = Copy(n.node1);
copy.node2 = Copy(n.node2);
}
return copy;
}
}
Node* Copy(Node* root) {
if (root == NULL)
return root;
std::unordered_map<Node*, Node*> completed;
std::deque<Node*> todo;
Node *ret = new Node(*scur);
completed.push_back(std::make_pair(root, ret));
todo.push_pack(root);
//while there's more nodes to duplicate
do {
//duplicate the node
Node* oldNode = todo.back();
Node* newNode = completed[cur];
todo.pop_back();
if(oldNode->left) {
auto iter = completed.find(oldNode->left);
//if it has a left child that needs duplicating, add it to the todo list
if (iter == completed.end()) {
newNode->left = new Node(*(oldNode->left));
completed.push_back(std::make_pair(oldNode->left, newNode->left));
todo.push_back(oldNode->left);
} else {
newNode->left = completed[oldNode->left];
}
}
if(oldNode->right) {
auto iter = completed.find(oldNode->right);
//if it has a right child that needs duplicating, add it to the todo list
if (iter == completed.end()) {
newNode->right = new Node(*(oldNode->right));
completed.push_back(std::make_pair(oldNode->right, newNode->right));
todo.push_back(oldNode->right);
} else {
newNode->right= completed[oldNode->right];
}
}
} while(todo.empty() == false)
//return the translation of the root
return ret;
}
Doesn't have stack overflow, root can be NULL, doesn't fail if left or right are NULL.
[Edit]Adam Rosenfield made me realize this was incorrect if there was loops in the network. Had to rewrite almost from scratch. Due to the large amount of code required, I prefer his code's for loop.
return new Node(*node);
Trick question?
You should write it recursively;
Node * Copy( Node * root )
{
Node * node_copy;
node_copy = new Node; // Assume Node1 and Node2 are initialized to 0
node_copy->content = root->content;
if( root->Node1 ) node_copy->Node1 = Copy( root->Node1 );
if( root->Node2 ) node_copy->Node2 = Copy( root->Node2 );
return node_copy;
}
So, this does not make any assumption on the data type
Given that a copy constructor exists that copies only the contents of a node and not its children:
Node* Copy(Node* root)
{
Node* copy = new Node(*root);
copy->left = Copy(root->left);
copy->right = Copy(root->right);
return copy;
}
In a more general sense, I would use copy-constructors that fully copy the entire data structure:
Node* Copy(Node* root)
{
return new Node(*root);
}