Balanced Search Trees

Definition:

Binary Search Trees with a worst-case height of O(log n) are called balanced trees

We can guarantee O(log n) performance for each search tree operation for balanced trees.

We will discuss one kinds of balanced trees:

AVL trees

AVL trees

Definition

An AVL tree is a binary search tree (BST) with an additional balance property, that is:

For any node in the tree, the height of the left subtree and the height of the right subtree differ by at most 1.

The balance property of a node

Balance information of a node t:

The balance information of a node t is given by:balance(t) = height(t.left)-height(t.right)

A balanced treeA binary search tree is balanced (in the AVL sense) if:for any node t in the tree, we have

balance(t)= 0, 1 or -1

Depth property of an AVL tree

The balance condition ensures a logarithmic depth.

About the proof

In order to show that an AVL tree has a logarithmic depth, we have to show that its height h has an upper bound that is:

logarithmic in n (n is the size of the tree = the number of nodes in the tree):

h ≤ logarithmic function of n.

In other words, we need to show that:n ≥ exponential function of h.

Therefore, we are looking for the minimum number of nodes n in an AVL tree of height h.

Theorem: the minimum number of nodes in an AVL tree

An AVL tree Th of height h has a minimum of fh+3 − 1 nodes where fh+3 is the (h + 3)th Fibonacci number.

Proof...

T h is of height h:

the height of one of the subtrees of the root of Th must be h − 1. The height of other subtree cannot be less than h − 2 (to maintain AVL balance property). Therefore,

size(Th) ≥ size(Sh−1) + size(Sh−2) + 1 for h ≥ 2 (1)

where Si is the subtree of Th with height i.

Insertion strategy in an AVL tree

• The tree must be balanced before any insertion.

• Insertion is performed the same way it is performed in a BST.

• After an insertion, only the nodes on the path from the root to the insertion point can have their balance altered.

• After a new node is inserted, we follow the path from the insertion point to the root and update the height (and balance) information of each node on that path.

• If a node that violates the balance property is found, the node is re-balanced by means of a series of operations called rotations.

• After a node is re-balanced and the height (or balance) information updated, no further re-balancing or updating is necessary in the remaining nodes in the path.

Possible situations after an insertion

• After a new node is inserted, we follow the path from the insertion point to the root and update the height (and balance) information of each node on that path.

• Say, a node X violating the balance property is found. There exist 4 possible situations:

Situation 1: the insertion is into the left sub-tree of the left child of X.

Situation 2: the insertion is into the right sub-tree of the left child of X.

Situation 3: the insertion is into the left sub-tree of the right child of X.

Situation 4: the insertion is into the right sub-tree of the right child of X.

Initial tree for situations 1 and 2

Situation 1

An insertion into the left sub-tree of the left child of X.

Situation 2

Insertion into the right sub-tree of the left child of X.

Initial tree for situations 3 and 4

Situation 3 (Symmetric to situation 2)

Insertion into the left sub-tree of the right child of X.

Situation 4 (Symmetric to situation 1)

Insertion into the right sub-tree if the right child of X.

In the following, we will consider the previous situations 2 by 2:

• The first and the fourth situations are symmetric. In these situations, the insertion is performed in the left sub-tree of the left child (situation 1) of X or in the right sub-tree of the right child of X (situation 4). We call these two cases: insertion from the outside.

• The second and the third situations are also symmetric. In these situations, the insertion is performed in the right sub-tree of the left child of X (situation 2) or in the left sub-tree of the right child of X (situation 3). We call these two cases: insertion from the inside.

Example: outside insertion

Consider the tree on the left. After inserting 1, the tree loses itsbalance.

We are in the situation of an outside insertion: 1 has been inserted to the left sub-tree of the left child of 8 (that lost its balance).

Example: inside insertion

Consider the tree on the left. After inserting 7, the tree loses its balance.

We are in the case of an inside insertion: 7 has been added tothe right sub-tree of the left child of 8 (that lost its balance)

Re-balancing the tree using rotations

We have two cases:

1 In the case of outside insertions: only a single rotation is necessary.

2 In the case of inside insertions: a double rotation is necessary.

Single rotations: LL

Situation 1: Insertion in the left sub-tree of the left child of the node that lost balance (here X is k2).Graphically:

LL Rotation(k2)1 k1 is the left child of k2.2 left child of k2 = right child of k1.3 left child of parent of k2 or right child of parent of k2 = k14 right child of k1 = k2.5 update height information for k2 and k1.6 return k1.

Pseudocode: LL rotation

Input: Binary Node k2

Procedure withLeftChild( k2) Set k1 k2.LeftIf (k2.parent != null)

if (k2 is Left child)k2.Parent.Left k1

elsek2.Parent.Right k1

k1.parent = k2.parent k2.parent = k1

k2.Left k1.Rightk1.Right k2// Also, update balance information!return k1

Example of an LL rotation

Single rotations: RR

Insertion in right sub-tree of right child of the node that lostbalance (Situation 4):Graphically:

RR Rotation(k2)1 k1 is the right child of k2.2 right child of k2 = left child of k1.3 left child of parent of k2 or right child of parent of k2 = k14 left child of k1 = k2.5 update height information for k2 and k1.6 return k1.

Pseudocode: RR rotation

Input: Binary Node k2

Procedure withRightChild( k2)Set k1 k2.RightIf (k2.parent != null)

if (k2 is Left child)k2.Parent.Left k1

elsek2.Parent.Right k1

k1.parent = k2.parent k2.parent = k1

k2.Right k1.Leftk1.Left k2// Update balance information too!return k1

Can we use the single rotation after an inside insertion?

No we cannot!!

Double rotations: LR

Insertion in the right sub-tree of the left child of the node thatlost balance (Situation 3).Graphically:

LR Rotation(k3)1 k1 is the left child of k32 k2 is the right child of k13 left child of k3 = right child of k24 right child of k1 = left child of k25 left child of parent of k3 or right child of parent of k3 = k26 left child of k2 = k17 right child of k2 = k38 update height information for k1, k3, and k29 return k2

Example of an LR rotation

K3

K1

K2

Double rotation: RL

Insertion in the left sub-tree of the right child of the node thatlost balance (Situation 2):

RL Rotation(k3)1 k1 is the right child of k32 k2 is the left child of k13 right child of k3 = left child of k24 left child of k1 = right child of k25 left child of parent of k3 or right child of parent of k3 = k26 left child of k2 = k37 right child of k2 = k18 update height information for k1, k3, k29 return k2

In fact a double rotation is a succession of to simple rotations.

This lead to the algorithm for double rotation:

Pseudocode: LR rotationInput: Binary Node k3Procedure doubleWithLeftChild( k3)

Set k3.LeftwithRightChild(k3.Left)return withLeftChild(k3)

Pseudocode: RL rotationInput: Binary Node k3Procedure doubleWithRightChild( k3)

Set k3.RightwithLeftChild(k3.Right)return withRightChild(k3)

Deletion strategy in an AVL tree

• Deletion is performed the same way it is performed in a BST.

• Deleting a node from an AVL tree may cause the tree to loose its balance: one of the nodes of the tree becomes unbalanced.

• After a deletion, only the nodes on the path from the root to the physically deleted node can have their balance altered.

• After a new node is deleted, we follow the path from the insertion point to the root and update the height (or balance) information of each node on that path.

• If a node that violates the balance property is found, the node is re-balanced by means of a series of rotations.

• After a node is re-balanced and the height (or balance) information updated, no further re-balancing or updating is necessary in the remaining nodes in the path.

Possible situations after a deletion

• After a node is deleted, we follow the path from the physically deleted node to the root and update the height (or balance) information of each node on that path.• Say, a node X violating the balance property is found.• There exist 6 possible situations. We will refer to these situations by:

Situation 1: deletion in the right sub-tree of X. The left child of X has a 0 balance. We refer to this situation by R0.Situation 2: deletion in the right sub-tree of X. The left child of X has balance 1. We refer to this situation by R1.Situation 3: deletion in the right sub-tree of X. The left child of X has balance -1. We refer to this situation by R-1.Situation 4 (symmetric to situation 1): deletion in the left sub-tree of X. The right child of X has a 0 balance. We refer to this situation by L0.Situation 5 (symmetric to situation 3): deletion in the left sub-tree of X. The right child of X has a 1 balance. We refer to this situation by L1.Situation 6 (symmetric to situation 2): deletion in the left sub-tree of X. The right child of X has a -1 balance. We refer to this situation by L-1.

Situation 1: R0

Deletion in the right sub-tree of X causes X to be unbalanced.The left child of X, has balance 0.

In a R0 situation, an LL rotation restores X’s balance.

Situation 1: R1

Deletion in the right sub-tree of X causes X to be unbalanced.The left child of X, has balance 1.

In a R1 situation, an LL rotation restores X’s balance.

Situation 1: R-1

Deletion in the right sub-tree of X causes X to be unbalanced.The left child of X, has balance -1.

In a R-1 situation, an LR rotation restores X’s balance.