A: DeCLarations GENerator: used to create the host language copy books for the table definitions. Also creates the DECLARE table. 

No Consider 2 trees below   Tree1 a b Tree 2 a b preorder = ab postorder = ba Preorder and postorder do not uniquely define a binary tree. Nor do preorder and level order (same example). Nor do postorder and level order.

A: Either by using the single row SELECT statements,or by using the CURSOR.

inorder = g d h b e i a f j c preorder = a b d g h e i c f j Scan the preorder left to right using the inorder sequence to separate left and right subtrees. For example, "a" is the root of the tree; "gdhbei" are in the left subtree; "fjc" are in the right subtree. "b" is the next root;

Here is some working C code... #include typedef struct node { int value; struct node *right; struct node *left; }mynode; mynode *root; mynode *add_node(int value); void levelOrderTraversal(mynode *root); mynode *closestAncestor(mynode* root, mynode* p, mynode* q); int main(int argc, char* argv[]) { mynode *node_pointers[7], *temp; root = NULL; // Create the BST. // Store the node

AVL trees are self-adjusting, height-balanced binary search trees and are named after the inventors: Adelson-Velskii and Landis. A balanced binary search tree has O(log n) height and hence O(log n) worst case search and insertion times. However, ordinary binary search trees have a bad worst case. When sorted data is inserted, the binary search tree is very unbalanced, essentially more of a linear list, with O(n) height and thus

Its 2^n - n So, if there are 10 nodes, you will have (1024 - 10) = 1014 different trees!! Confirm it yourself with a small number if you dont believe the formula.    

if((num & x00001000)==x00001000)   Note that the digits represented here are in hex. 0 0 0 0 1 0 0 0 ^ | x0000 0000 0000 0000 0001 0000 0000 0000 = 32 bits ^ ^ ^ | | | 0th bit 20th bit 32nd bit

Here are some ways to reverse the bits in an integer. Method1 unsigned int num; // Reverse the bits in this number. unsigned int temp = num; // temp will have the reversed bits of num. int i; for (i = (sizeof(num)*8-1); i; i--) { temp = temp | (num & 1); temp = 1; } temp = temp | (num

The AND operator Truth Table ----------- 0 AND 0 = 0 0 AND 1 = 0 1 AND 0 = 0 1 AND 1 = 1 x AND 0 = 0 x AND 1 = x We use bitwise "and" to test if certain bit(s) are one or not. And'ing a value against a pattern with ones only in the

This is one of the most frequently asked interview questions of all times... There are a number of ways to count the number of bits set in an integer. Here are some C programs to do the same. Method1 This is the most basic way of doing it. #include int main() { unsinged int num=10; int ctr=0; for(;num!=0;num>>=1) {

Since traversing the three is the most frequent operation, a method must be devised to improve the speed. This is where Threaded tree comes into picture. If the right link of a node in a tree is NULL, it can be replaced by the address of its inorder successor. An extra field called the rthread is used. If rthread is equal to 1, then it means that the right

Use Geometric progression. M + (N ^ (n-1)) = (1 - (N ^ n)) / (1 - N) Here (N ^ (n-1)) is the number of leaf-nodes. Solving for this leads to the answer Leaf nodes = M * (N - 1) + 1   Suppose you have a 3-ary tree A B C D E F G H I J K L M   So,

This is one of the most frequently asked interview questions of all times... There are a number of ways to count the number of bits set in an integer. Here are some C programs to do the same. Method1 This is the most basic way of doing it. #include int main() { unsinged int num=10; int ctr=0; for(;num!=0;num>>=1) {

Depth first search (DFS) Depth First Search (DFS) is a generalization of the preorder traversal. Starting at some arbitrarily chosen vertex v, we mark v so that we know we've visited it, process v, and then recursively traverse all unmarked vertices adjacent to v (v will be a different vertex with every new method call). When we visit a vertex in which all of its neighbors have been