Problem Analysis for Zeke and his Password Encryption

PROBLEM: Zeke and his Password Encryption

This problem can be solved as follows:
This problem can be solved easily using the concept of AVL trees.
Output is preorder traversal of the AVL tree created after balancing.

Problem setter's solution in C:

#include<stdio.h>
#include<stdlib.h>

struct node
{
    int key;
    struct node *left;
    struct node *right;
    int height;
};

int height(struct node *N)
{
    if (N == NULL)
        return 0;
    return N->height;
}

int max(int a, int b)
{
    return (a > b)? a : b;
}

struct node* newNode(int key)
{
    struct node* node = (struct node*)malloc(sizeof(struct node));
    node->key   = key;
    node->left   = NULL;
    node->right  = NULL;
    node->height = 1;
    return(node);
}

struct node *rightRotate(struct node *y)
{
    struct node *x = y->left;
    struct 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.
struct node *leftRotate(struct node *x)
{
    struct node *y = x->right;
    struct 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(struct node *N)
{
    if (N == NULL)
        return 0;
    return height(N->left) - height(N->right);
}

struct node* insert(struct node* node, int key)
{
    /* 1.  Perform the normal BST rotation */
    if (node == NULL)
        return(newNode(key));

    if (key < node->key)
        node->left  = insert(node->left, key);
    else
        node->right = insert(node->right, key);

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

    /* 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 node;
}

void preOrder(struct node *root)
{
    if(root != NULL)
    {
        printf("%d", root->key);
        preOrder(root->left);
        preOrder(root->right);
    }
}

int main()
{
  struct node *root = NULL;
  int n,a,i,t;
  scanf("%d",&t);
  while(t--)
  {
  scanf("%d",&n);
  for(i=0;i<n;i++)
  {
    scanf("%d",&a);
    root=insert(root,a);
  }
  preOrder(root);
  root=NULL;
  printf("\n");
  }
  return 0;
}