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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; }
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.
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); }
How do you recursively reverse a singly linked list using c plus plus?
In this case recursion is not necessary, an iterative process is more efficient. Start by pointing at the head node. While this node has a next node, detach its next node and insert that node at the head. Repeat until the original head node has no next node. At that point the head has become the tail and all the nodes are completely reversed. The following example shows how this can be implemented, where the list object contains a head node (which may be NULL), and each node has a next node. The tail node's next node is always NULL. void reverse(list& lst) { if( node* p=lst.head ) { while(p->next) { node* n=p.next; // point to the next node p.next=n.next; // detach the next node n.next=lst.head; // insert the detached node at the head lst.head=n; // set the new head node } } }
If the degree of a node is 0 then the tree is called?
This would be just a single node, since no edges (you can think of degree as the number of edges connected to a node). If you are talking about the in-degree, or out-degree of a node being zero, this can happen many times in a directed graph (in-degree = # edges going IN to node, out-degree = # edges going out...).
Difference between in-degree and out-degree of a vertex in data structures?
In-degree is a count of the number of ties directed to the node, and out-degree is the number of ties that the node directs to others.
Pacemakers may be useful in treating?
a third degree AV node block
Why don't we allow a minimum degree of B-Tree is 1?
One important property of a B-Tree is that every node except for the root node must have at least t-1 keys where t is the minimum degree of the B tree. With t-1 keys, each internal node will have at least t children [Cormen et al., Introduction To Algorithms Third Edition, MIT Press, 2009 p. 489].If we allow a minimum degree of 1, then each internal node will have 1-1=0 keys!
What is strictly binary tree?
If every non-terminal node (any node except root node whose degree is not zero) in a binary tree consists of non-empty left and right subtree, then such a tree is called strictly binary tree.
Id like to know what is to long to consider for a 1st degree block .22 to?
First degree atrio-ventricular node block is a PR segment longer than 0.2 seconds.
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; } }
How would the ECG be affected If there is a complete block between the SA node and the AV node?
If the atrial conduction system (the SA node and bachmann bundles) and the ventricular conduction system (the AV node and purkinje system) are completely separated, third degree heart block develops. In this case, the SA node will continue to depolarize the atria at its own intrinsic rate and the ventricles will be depolarized by a focus either in the AV node or in the ventricles.
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;
