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Q: Is null node equal to leaf node?

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In this representation, each node contains two pointers, one pointing to its parent (null in the case of root node) and the other pointing to its child node (null in the case of leaf nodes).

#include <stdio.h> #include <stdlib.h> /* A binary tree node has data, pointer to left child and a pointer to right child */ struct node { int data; struct node* left; struct node* right; }; /* Function to get the count of leaf nodes in a binary tree*/ unsigned int getLeafCount(struct node* node) { if(node NULL && node->right==NULL) return 1; else return getLeafCount(node->left)+ getLeafCount(node->right); } /* Helper function that allocates a new node with the given data and NULL left and right pointers. */ struct node* newNode(int data) { struct node* node = (struct node*) malloc(sizeof(struct node)); node->data = data; node->left = NULL; node->right = NULL; return(node); } /*Driver program to test above functions*/ int main() { /*create a tree*/ struct node *root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); /*get leaf count of the above created tree*/ printf("Leaf count of the tree is %d", getLeafCount(root)); getchar(); return 0; }

for (node=head; node!=null; node=node->next) printnode(node);

int countleaves(struct node* root){ if(root!=null) { countleaves(root->left); if(root->left==NULL&&root->right==NULL) { count++; } countleaves(root->right); } }

height and depth of a tree is equal... but height and depth of a node is not equal because... the height is calculated by traversing from leaf to the given node depth is calculated from traversal from root to the given node.....

Induction: 1. A tree of one node has two NULL-pointers. 2. Whenever you add a node to a tree, you remove one NULL-pointer (from the parent of the new node), and add two (the child's of the new node) in the same time.

struct node{ int data; struct node *left, *right; }; typedef struct node node; non recursive method : main() { node * root = NULL , *new = NULL *temp1 =NULL , * temp2 = NULL; int num =1; printf(" Enter the elements of the tree( enter 0 to exit)\n"); while(1) { scanf("%d", &num); if(num==0) break; new = malloc(sizeof(node)); new->left = new->right = NULL; new->data = num; if( root NULL) root->left = new; else insert( new,root->left); } }

There must be no NULL links in a linked list except in the tail node (because nothing comes after the tail). In doubly-linked lists, the head node also has a NULL link because nothing comes before the head. Every node in a list must be reachable from the previous (or next) node in the list, except the head node which is the only node we must keep track of. A NULL link would render all nodes that follow it unreachable, resulting in a resource leak -- unless we take steps to keep track of the node that would have come next before breaking the link to it. That node becomes the head node of another list.

A null tree.

Queues are a first in first out structure (FIFO). This means all extractions occur at the head of the list and all insertions occur at the tail. This requires that you maintain pointers to the head and tail node to allow constant time insertion and extraction of nodes. The nodes themselves are singly linked, each pointing to the next node. The tail node always points to NULL until a new node is insert, which the tail points to. The new node then becomes the tail. When extracting a head node, its next node (which may be NULL if it is also the last node) becomes the new head of the list. If the head is NULL, the tail is also set to NULL.

Since a binary search tree is ordered to start with, to find the largest node simply traverse the tree from the root, choosing only the right node (assuming right is greater and left is less) until you reach a node with no right node. You will then be at the largest node.for (node=root; node!= NULL&&node->right != NULL; node=node->right);This loop will do this. There is no body because all the work is done in the control expressions.

void delete_last (node **head) { node *tmp = *head; node *prev = NULL; if (!tmp) { } else if (!tmp->next) { free(tmp); *head = NULL; } else { while (tmp->next) { prev = tmp; tmp = tmp->next; } prev->next = NULL; free(tmp); } return; }

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