ADVANCED DATA STRUCTURES PROGRAM • LEVEL 14 — BALANCED BINARY SEARCH TREES

Demonstrate All Four AVL Rotation Cases

Learn how to demonstrate all four avl rotation cases using a clear C program.

AdvancedSelf-balancing BSTLogarithmic operations

PROBLEM UNDERSTANDING

Input and expected output

Sample input
No input required
Sample output
20 20 20 20

COMPLETE C PROGRAM

Complete C implementation

ads-avl-four-rotation-cases.c
Open in compiler
#include <stdio.h>
#include <stdlib.h>

struct Node { int key, height; struct Node *left, *right; };
int height(struct Node *node) { return node ? node->height : 0; }
int maximum(int a, int b) { return a > b ? a : b; }
struct Node *new_node(int key) { struct Node *node = malloc(sizeof *node); if (!node) exit(EXIT_FAILURE); node->key = key; node->height = 1; node->left = node->right = NULL; return node; }
void update(struct Node *node) { node->height = 1 + maximum(height(node->left), height(node->right)); }
struct Node *rotate_right(struct Node *root) { struct Node *pivot = root->left, *middle = pivot->right; pivot->right = root; root->left = middle; update(root); update(pivot); return pivot; }
struct Node *rotate_left(struct Node *root) { struct Node *pivot = root->right, *middle = pivot->left; pivot->left = root; root->right = middle; update(root); update(pivot); return pivot; }
struct Node *balance_node(struct Node *root)
{
    update(root); int balance = height(root->left) - height(root->right);
    if (balance > 1) { if (height(root->left->left) < height(root->left->right)) root->left = rotate_left(root->left); return rotate_right(root); }
    if (balance < -1) { if (height(root->right->right) < height(root->right->left)) root->right = rotate_right(root->right); return rotate_left(root); }
    return root;
}
struct Node *insert(struct Node *root, int key) { if (!root) return new_node(key); if (key < root->key) root->left = insert(root->left, key); else if (key > root->key) root->right = insert(root->right, key); else return root; return balance_node(root); }
struct Node *minimum_node(struct Node *root) { while (root->left) root = root->left; return root; }
struct Node *delete_key(struct Node *root, int key)
{
    if (!root) return NULL;
    if (key < root->key) root->left = delete_key(root->left, key);
    else if (key > root->key) root->right = delete_key(root->right, key);
    else if (!root->left || !root->right) { struct Node *child = root->left ? root->left : root->right; free(root); return child; }
    else { struct Node *next = minimum_node(root->right); root->key = next->key; root->right = delete_key(root->right, next->key); }
    return balance_node(root);
}
void inorder(struct Node *root) { if (root) { inorder(root->left); printf("%d ", root->key); inorder(root->right); } }
void free_tree(struct Node *root) { if (root) { free_tree(root->left); free_tree(root->right); free(root); } }

int main(void)
{
    int cases[4][3] = {{30,20,10},{10,20,30},{30,10,20},{10,30,20}};
    for (int row = 0; row < 4; row++) { struct Node *root = NULL; for (int i = 0; i < 3; i++) root = insert(root, cases[row][i]); printf("%d%c", root->key, row == 3 ? '\n' : ' '); free_tree(root); }
    return 0;
}

GUIDED CODE TOUR • NOT LIVE EXECUTION

Study the program line by line

Use the real compiler button above to run and debug with different inputs.

CURRENT STEP

Select Start to walk through the important lines.

SELECTED LINE

No line selected

EXPECTED OUTPUT FOR THE SAMPLE

20 20 20 20
0%Step 0 of 0

PROGRAM EXPLANATION

Algorithm and explanation

  1. Read the required input values.
  2. Trigger LL, RR, LR and RL imbalance patterns and observe their common balanced root.
  3. Display the computed result.

Trigger LL, RR, LR and RL imbalance patterns and observe their common balanced root.

EFFICIENCY

Time and space complexity

Time complexity

O(log n) per insertion

Auxiliary space

O(h)

DEBUGGING CHECKLIST

Common mistakes

Check this

Use the correct format specifier for every variable.

Check this

Initialize variables before using their values.

Check this

Check braces, semicolons and input order carefully.

Try it yourself

Practice: Run the program with the sample input, predict its output, and then test one boundary case of your own.