Welcome to Lesson 14!

Learning Objectives
By the end of this class, you should know...

  • What is an AVL Tree?
  • What is the definition of a balanced tree according to the AVL tree property?
  • How do AVL trees maintain their balance?
  • What is a simple left rotation and when do you use it?
  • What is a simple right rotation and when do you use it?
  • How do you write the pseudocode for simple left and right rotations?
  • What is a double rotation?
  • What is a left-right rotation and when do you use it?
  • What is a right-left rotation and when do you use it?

  • Midterm 2 on Thursday
  • Return Quizzes
  • Collect Lab 7
  • Lab 8 assigned
  • Project Report 3 (one per group) and Peer Evaluation (each person to submit) due Thursday.
  • Last day to drop with a "W" is this Friday

Review Activity

  • What is the max heap property?
  • Trace build heap on the following values: 10, 6, 4, 9, 12, 5, 3
  • Trace heap insert on the heap created above and the value 15.


  • Exchange the right most bottom key with the root
  • Rebuild the Heap
  • Repeat



    n = length(A)

    for (i = n down to 2)

        A[1] <--> A[i] //swap

        Heap_size(A)-- //consider your heap to be one smaller

        Max-Heapify(A,1) //restore max heap property

Big O Run-Time: O(n log n)

Group Activity: Heap Sort Practice

  • Take out your heap handout from last class and try the heap-sort algorithm question with a partner.

AVL Trees

  • An AVL tree (Georgy Adelson-Velsky and Evgenii Landis tree) is a self-balancing BST.

AVL Tree Property

  • Requires that the heights of the left and right children of every node differ by at most one.
|left.height - right.height| <= 1
  • Maintains a balanced tree to achieve O(log n) time for ALL operations.
    • Review: What is the Big-O runtime of BST operations in best, worst and average cases?

AVL Tree Insert

  • We need to insert nodes in such a way that the AVL tree property is satisfied.
  • Thus, inside of our BST Nodes we are going to keep track of the height of the nodes.

struct Node
    bstdata data;

    int height;

    Node* left;
    Node* right;

    Node(): left(NULL), right(NULL), height(0){}
    Node(bstdata newdata): left(NULL), right(NULL), data(newdata), height(0){}

  • Insertion has 2 steps:
  1. Call BST insert
  2. Rebalance the Tree
  • To balance the tree we use what are called "rotations" to alter the location of certain nodes.

AVL Insert Pseudocode (Partial)

Node avlInsert(Node root, int value)
    //BST style insertion
if (root == NULL)
    return new Node(value)
else if(root.data >= value)
    root->left = avlInsert(root->left, value)
    root->right = avlInsert(root->right, value)
//value has now been inserted
//Is the resulting tree unbalanced?
int balance = root->left->height - root->right->height
if (balance > 1 || balance < -1)
    //re-balance the tree using rotations
    //to be filled in later in class



  • With rotations, we want the height of the tree to become shorter (tree looks less like a linked list)
  • Important: These rotations maintain the order of the BST: T1 < A < T2 < B < T3.

Simple Examples

Tree rotation.png

Image source.

Right Rotation

        Before Rotation        After Rotation

  • Note that the upper case letters represent nodes in the tree, while the lower case letters represent possible subtrees.
  • What happened to subtree c during the rotation?

Example Right Rotation

Right Rotation Pseudocode

Nodeptr Right_Rotate(Nodeptr root)
Nodeptr newRoot = root->left
root->left = newRoot->right
newRoot->right = root
root->height = maxHeight(root->left, root->right) + 1
newRoot->height = maxHeight(newRoot->left, newRoot->right) + 1
return newRoot

Left Rotation 

                    Before Left Rotation                               After Left Rotation

Example Left Rotation

Activity 14.1: Left Rotation Pseudocode

  • Now write the pseudocode for left rotation.
  • Use the images and the right rotation pseudocode to help yourself.

Activity 14.2: Single Rotation Practice

  • Trace the following AVL insertions:
  • Example 1: Insert 1, 2, 3
  • Example 2: Insert 2, 1, 4, 6, 3, 5
  • Example 3: Insert 2, 1, 4, 6, 7
  • Example 4: Insert 2, 1, 4, 3, 6, 7

Double Rotations

  • Sometimes a single rotation is not enough to balance a tree.
  • Consider the following example:
  • Our reaction might be to right rotate it, like so:
  • However, we are now no better off than when we started.
  • In cases like these (zig-zag pattern), we need to use a double rotation.
  • We will need to do either a left-right rotation or a right-left rotation, depending on the situation.
  • If the left subree is "right heavy" (i.e. left.height - right.height < -1)
    • rotate left then rotate right
  • If the right subtree is "left heavy" (i.e. left.height - right.height > 1)
    • rotate right then rotate left
  • In the example above, we have a left heavy right subtree.
  • Therefore, we will do a left-right rotation.

Step 1: Rotate left around the root.left, i.e. node A (to get a familar shape)

Step 2: Rotate right around the root, i.e. B (to get a balanced tree)


Example of left-right rotation

Image source.
  • If we have the opposite situation, where the right subtree is left heavy:
  • Then, we must perform a right-left rotation.

Step 1: Rotate right around root.right, i.e. A (to get a familar shape)

Step 2: Rotate left around root, i.e. B (to get a balanced tree)


Example of right-left rotation

Image source.

Double Rotation Videos

Activity 14.3: Double Rotation Practice

  • Trace the following AVL insertions:
    • 3 2 1 4 5 6 7 16 15 14
  • When you are finished, double check your work at the bottom.

AVL Insert Pseudocode (Full)

Nodeptr avlInsert(Nodeptr root, int value)
    //BST style insertion
    if (root == NULL)
    if (value < root.data)
        root->left  = avlInsert(root->left, value);
        root->right = avlInsert(root->right, value);
    //Check the balance
    int balance = height(root->left) - height(root->right);
   //Check for 4 cases

    // Simple Right Rotation
    if (balance > 1 && value < root->left->data)
        return rightRotate(root);
    // Simple Left Rotation
    else if (balance < -1 && value > root->right->data)
        return leftRotate(root);
    // Left Right Case
    else if (balance > 1 && value > root->left->value)
        root->left =  leftRotate(root->left);
        return rightRotate(root);
    // Right Left Case
    else if (balance < -1 && value < root->right->data)
        root->right = rightRotate(root->right);
        return leftRotate(root);
    //return the node pointer
    return root;

Double Insertion Solution

Image source

Wrap Up

  • With your partner, answer the learning objective questions from the beginning of the lesson.

Upcoming Assignments

  • Lab 8 due one week from today