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How can schmorls node be treated?
Schmorl's nodes are lesions in which disc material makes its way into the vertebral body. They can usually be treated with pain killers and non-surgical methods. Sometimes however, the level of pain caused by them is so severe that fusion surgery is performed.
Write an iterative function to search an element in a binary search tree?
_node* search (_node* head, _key key) { _node* node; for (node=head; node != NULL;;) { if (key == node->key) return node; else if (key < node.>key) node = node->left; else node = node->right; } return node; }
How to Print data of all nodes of linked list?
for (node=head; node!=null; node=node->next) printnode(node);
How to find the mirror image of a binary tree?
Refer to http://cslibrary.stanford.edu/110/BinaryTrees.html void mirror(struct node* node) { if (node==NULL) { return; } else { struct node* temp; // do the subtrees mirror(node->left); mirror(node->right); // swap the pointers in this node temp = node->left; node->left = node->right; node->right = temp; } }
Can we use doubly linked list as a circular linked list?
Yes. The tail node's next node is the head node, while the head node's previous node is the tail node.
Algoritm for deleting the last element from a list?
Given a list and a node to delete, use the following algorithm: // Are we deleting the head node? if (node == list.head) { // Yes -- assign its next node as the new head list.head = node.next } else // The node is not the head node { // Point to the head node prev = list.head // Traverse the list to locate the node that comes immediately before the one we want to delete while (prev.next != node) { prev = prev.next; } end while // Assign the node's next node to the previous node's next node prev.next = node.next; } end if // Before deleting the node, reset its next node node.next = null; // Now delete the node. delete node;
Is null node equal to leaf node?
No. A leaf node is a node that has no child nodes. A null node is a node pointer that points to the null address (address zero). Since a leaf node has no children, its child nodes are null nodes.
What is an intrathoracic node?
An intrathoracic node is a node within the chest cavity.
How many pointers will have to be changed if a node is deleted from a linear linked list?
For a singly-linked list, only one pointer must be changed. If the node about to be deleted (let's call it node for the sake of argument) is the head of the list, then the head node pointer must be changed to node->next. Otherwise, the node that comes before the deleted node must change its next pointer to node->next. Note that given a singly-linked node has no knowledge of its previous node, we must traverse the list from the head in order to locate that particular node, unless the node is the head of the list: void remove (List* list, Node* node) { if (!list !node) return; // sanity check!if (list->head == node) {list->head = node->next;} else {Node* prev = list->head;while (prev->next != node) prev = prev->next; // locate the node's previous nodeprev->next = node->next;}} Note that the remove function only removes the node from the list, it does not delete it. This allows us to restore the node to its original position, because the node itself was never modified (and thus still refers to its next node in the list). So long as we restore all removed nodes in the reverse order they were removed, we can easily restore the list. In order to delete a node completely, we simply remove it and then free it:void delete (List* list, Node* node) {if (!list !node) return; // sanity check!remove (list, node);free (node);} For a doubly-linked list, either two or four pointers must be changed. If the node about to be deleted is the head node, then the head node pointer must be changed to n->next and n->next->prev must be changed to NULL, otherwise, n->prev->next becomes n->next. In addition, if the node about to be deleted is the tail node, then the tail node pointer must be changed to n->prev and n->prev->next must be changed to NULL, otherwise, n->next->prev becomes n->prev. Deletion from a doubly-linked list is generally quicker than deletion from a singly linked list because a node in a doubly-linked list knows both its previous node and its next node, so there's no need to traverse the list to locate the previous node to the one being deleted. void remove (List* list, Node* node) {if (!list !node) return; // sanity check!if (list->head == node) {list->head = node->next;node->next->prev = NULL;} else {node->prev->next = node->next; }if (list->tail == node) {list->tail = node->prev;node->prev->next = NULL;} else {node->next->prev = node->prev; }} Again, to physically delete the node we simply remove and then free the node:void delete (List* list, Node* node) {if (!list !node) return; // sanity check!remove (list, node); free (node); }
What is Cardiopulmonary node?
cardiopulmonary node
What is a simple node?
In computer programming, a simple node is a structure that holds two pointers. The first pointer refers to the data that is indirectly associated with the node, while the other refers to the next node in the sequence. Simple nodes are typically used in forward lists (singly-linked lists) where we keep track of the head node only. The last node in the list refers to NULL as there can be no next node after the last node. We can insert new data into the list knowing only the address of the data. A new node is constructed which then refers to the data and to the current head of the list. The new node then becomes the head of the list. Insertions at the head are performed in constant time. Inserting anywhere else in the list requires that we traverse the nodes recursively, following each node's next pointer, starting from the head, stopping at the node that should come after the new node. That node then becomes the new node's next node (if the node reaches the end of the list, the next node is NULL). Since we traversed recursively, we can return the newly inserted node to the caller, which can then be assigned to the caller's next node. All other recursions simply return the node we traversed to, thus leaving the caller's next node unchanged. The following example shows a simple node implementation for an ordered list of type int: typedef struct Node_t { // a simple node int* data; Node_t* next; } Node; Node* insert (int* data, Node* node) { // insert data if (!node *node->data > *data ) { // insertion point reached (end recursions) Node* pnode = malloc (sizeof (Node)); // create a new node pnode->data = data; // attach the data pnode->next = node; // refer to the next node (may be NULL) return pnode; // return the new node so the previous node can be updated } // if we get this far, the given node comes before the new data // delegate to the next node, the return value is the new or existing next node node->next = insert (data, node->next); return node; // return the node (it remains the previous node's next node) } void print_all (Node* node) { // print the given node and all that follow it while (node) { printf ("%d ", *node->data); node = node->next; } } Node* destroy_all (Node* node) // destroy the given node and all that follow it while (node) { Node* next = node->next; // refer to the next node (may be NULL) free (node->data); // release the current node's data free (node); // release the node itself node = next; // make the next node current (may be NULL) } return node; // always NULL } int main(void) { // The test program... srand ((unsigned) time (NULL)); // seed a random number generator Node* head = NULL; // the head node of the list (initially NULL for an empty list) for (int i=0; i<10; ++i) { // insert 10 random integers int* pint = malloc (sizeof (int)); // allocate a new integer on the free store if (!pint) break; // allocation failed, stop inserting! *pint = rand () % 10; // assign a random integer value (range 0:9) head = insert (pint, head); // insert the data (the return value is the new or existing head) } print_all (head); // print all the nodes starting from the head node head = delete_all (head); // destroy all the nodes starting from the head (returns NULL) return 0; }
What pacemaker of the heart initiates depolarization under normal conditions?
The primary pacemaker of the mammalian heart is the sino-atrial node. If the SA node fails, the atrioventricular node (AV node) takes over pacemaking.
