Question

PRIORITY QUEUE USING AVL TREE [ c++ ]

Implement a PriorityQueue where we will store the underlying data as an AVL Tree. For each of the fundamental operations of a PriorityQueue, explain briefly how you would implement it with an AVL Tree. If something is a standard AVL tree operation, such as a re-balance, you need only refer to it. The operations are:

1. insertion

2. extractMin()

3.min()

4.extractMax()

5.Max()

Answer 1

The operations:-

1> INSERTION:-

To make sure that the given tree remains AVL after every insertion, you must augment the standard BST insert operation to perform some re-balancing. Following are two basic operations that can be performed to re-balance a BST without violating the BST property (keys(left) < key(root) < keys(right)).
1) Left Rotation
2) Right Rotation

T1, T2 and T3 are subtrees of the tree 
rooted with y (on the left side) or x (on 
the right side)           
     y                               x
    / \     Right Rotation          /  \
   x   T3   - - - - - - - >        T1   y 
  / \       < - - - - - - -            / \
 T1  T2     Left Rotation            T2  T3
Keys in both of the above trees follow the 
following order 
 keys(T1) < key(x) < keys(T2) < key(y) < keys(T3)
So BST property is not violated anywhere.

Steps to follow for insertion
Let the newly inserted node be w
1) Perform standard BST insert for w.
2) Starting from w, travel up and find the first unbalanced node. Let z be the first unbalanced node, y be the child of z that comes on the path from w to z and x be the grandchild of z that comes on the path from w to z.
3) Re-balance the tree by performing appropriate rotations on the subtree rooted with z. There can be 4 possible cases that need to be handled as x, y and z can be arranged in 4 ways. Following are the possible 4 arrangements:
a) y is left child of z and x is left child of y (Left Left Case)
b) y is left child of z and x is the right child of y (Left-Right Case)
c) y is the right child of z and x is the right child of y (Right Right Case)
d) y is the right child of z and x is left child of y (Right Left Case)

Following are the operations to be performed in above mentioned 4 cases. In all of the cases, we only need to re-balance the subtree rooted with z and the complete tree becomes balanced as the height of subtree (After appropriate rotations) rooted with z becomes the same as it was before insertion.

2> EXTRA MINI():-

this basically return you the least valued element in an AVL tree

3> MIN():-

this method returns the least valued element in the corresponding segment of the binary tree

4>EXTRA MAX():-

this basically returns you the highest valued element in an AVL tree.

5> MAN():-

this method returns the highest valued element in the corresponding segment of the binary tree.

here below is a code that may help you to understand the creation of AVL tree.---

<------------------------------------------code start-------------------------------------------->

// C++ program to insert a node in AVL tree
#include<bits/stdc++.h>
using namespace std;

// An AVL tree node
class Node
{
   public:
   int key;
   Node *left;
   Node *right;
   int height;
};

// A utility function to get maximum
// of two integers
int max(int a, int b);

// A utility function to get the
// height of the tree
int height(Node *N)
{
   if (N == NULL)
       return 0;
   return N->height;
}

// A utility function to get maximum
// of two integers
int max(int a, int b)
{
   return (a > b)? a : b;
}

/* Helper function that allocates a
new node with the given key and
NULL left and right pointers. */
Node* newNode(int key)
{
   Node* node = new Node();
   node->key = key;
   node->left = NULL;
   node->right = NULL;
   node->height = 1; // new node is initially
                   // added at leaf
   return(node);
}

// A utility function to right
// rotate subtree rooted with y
// See the diagram given above.
Node *rightRotate(Node *y)
{
   Node *x = y->left;
   Node *T2 = x->right;

   // Perform rotation
   x->right = y;
   y->left = T2;

   // Update heights
   y->height = max(height(y->left),
                   height(y->right)) + 1;
   x->height = max(height(x->left),
                   height(x->right)) + 1;

   // Return new root
   return x;
}

// A utility function to left
// rotate subtree rooted with x
// See the diagram given above.
Node *leftRotate(Node *x)
{
   Node *y = x->right;
   Node *T2 = y->left;

   // Perform rotation
   y->left = x;
   x->right = T2;

   // Update heights
   x->height = max(height(x->left),     
                   height(x->right)) + 1;
   y->height = max(height(y->left),
                   height(y->right)) + 1;

   // Return new root
   return y;
}

// Get Balance factor of node N
int getBalance(Node *N)
{
   if (N == NULL)
       return 0;
   return height(N->left) - height(N->right);
}

// Recursive function to insert a key
// in the subtree rooted with node and
// returns the new root of the subtree.
Node* insert(Node* node, int key)
{
   /* 1. Perform the normal BST insertion */
   if (node == NULL)
       return(newNode(key));

   if (key < node->key)
       node->left = insert(node->left, key);
   else if (key > node->key)
       node->right = insert(node->right, key);
   else // Equal keys are not allowed in BST
       return node;

   /* 2. Update height of this ancestor node */
   node->height = 1 + max(height(node->left),
                       height(node->right));

   /* 3. Get the balance factor of this ancestor
       node to check whether this node became
       unbalanced */
   int balance = getBalance(node);

   // If this node becomes unbalanced, then
   // there are 4 cases

   // Left Left Case
   if (balance > 1 && key < node->left->key)
       return rightRotate(node);

   // Right Right Case
   if (balance < -1 && key > node->right->key)
       return leftRotate(node);

   // Left Right Case
   if (balance > 1 && key > node->left->key)
   {
       node->left = leftRotate(node->left);
       return rightRotate(node);
   }

   // Right Left Case
   if (balance < -1 && key < node->right->key)
   {
       node->right = rightRotate(node->right);
       return leftRotate(node);
   }

   /* return the (unchanged) node pointer */
   return node;
}

// A utility function to print preorder
// traversal of the tree.
// The function also prints height
// of every node
void preOrder(Node *root)
{
   if(root != NULL)
   {
       cout << root->key << " ";
       preOrder(root->left);
       preOrder(root->right);
   }
}

// Driver Code
int main()
{
   Node *root = NULL;
  
   /* Constructing tree given in
   the above figure */
   root = insert(root, 10);
   root = insert(root, 20);
   root = insert(root, 30);
   root = insert(root, 40);
   root = insert(root, 50);
   root = insert(root, 25);
  
   /* The constructed AVL Tree would be
               30
           / \
           20 40
           / \ \
       10 25 50
   */
   cout << "Preorder traversal of the "
           "constructed AVL tree is \n";
   preOrder(root);
  
   return 0;
}

<------------------------------------------code end-------------------------------------------->