binary trees -2 chapter 6 1. 2 threaded trees (depth first) binary trees have a lot of wasted space:...

1

Binary trees -2

Chapter 6

3

Threaded trees (depth first)

• Binary trees have a lot of wasted space: the leaf nodes each have 2 null pointers

• We can use these pointers to help us in inorder traversals

• We have the pointers reference the next node in an inorder traversal; called threads

• We need to know if a pointer is an actual link or a thread, so we keep a Boolean for each pointer

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Threaded Tree Example

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75

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Threaded Tree Traversal

• Start at the leftmost node in the tree, print it, and follow its right thread– following a thread to the right, will output

the node and continue to its right– following a link to the right, will go to the

leftmost node, print it, and continue

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Threaded Tree Traversal -1

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Start at leftmost node, print it

Output1

7


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Follow thread to right, print node

Output13

8


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Follow link to right, go to leftmost node and print

Output135

9


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Output1356

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Output13567

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Threaded Tree Traversal - 6

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Output135678

8/8/02 12Amir Kamil

Threaded Tree Traversal - 7

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Output1356789

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Output135678911

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Output13567891113

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Threaded Tree Traversal CodeNode leftMost(Node n) { Node ans = n; if (ans == null) { return null; } while (ans.left != null) { ans = ans.left; } return ans;}

void inOrder(Node n) { Node cur = leftmost(n); while (cur != null) { print(cur); if (cur.rightThread) { cur = cur.right; } else { cur = leftmost(cur.right); } }}

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Traversal Through Tree Transformation (read only)

• Systematic visiting of all nodes in the tree.

• Each node visited once, does not specify order.

• Breadth-first traversal: visit each node, starting from the lowest level and moving down by level, visiting nodes from left to right.

• Depth-first traversals: go in subtree as deep as you can, backtrack. Differ on the order of visiting root, left, right subtrees.

• preorder: VLR.

• inorder: LVR.

• postorder: LRV.

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Insertion

• searching does not modify the tree.• To insert a new node with key el(new), a tree node with a

dead end has to be reached, new node attached to it.• found using same procedure as searching: compare key

of currently scanned node el(cur) to el(new).

• If el(new) less than the key el(cur) try left child; otherwise try right child.

• If the child is empty, discontinue search and make the child point to a new node of key el.

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Inserting in threaded tree

•for inorder, only takes care of successors.•Node with a right child: has its successor somewhere in the right subtree, does not need a thread.•Why ? Threads are for "climbing up the tree", not for going down.•A node with no right child has its successor somewhere. Inherits successor from parent.•If a node becomes a left node, its parent is successor.

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Deleting (try)

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Balancing a tree

• The main idea of DSW algorithm is a rotation• There are two types of rotation, left and right• Rotate Right( Gr, Par, Ch )

– If Par is not the root of the tree• Grandparent Gr of child Ch, becomes Ch’s parent by replacing Par;

– Right subtree of Ch becomes left subtree of Ch’s parent Par;

– Node Ch acquires Par as its right child

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A picture will help

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AVL tree

• The height of left and right subtree of every node differ by at most one.

AVL Trees 27

Balanced and unbalanced BST

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2 5

1 3

1

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7

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2 6

5 71 3

Is this “balanced”?

AVL Trees 28

Perfect Balance

• Want a complete tree after every operation– tree is full except possibly in the lower right

• This is expensive– For example, insert 2 in the tree on the left and

then rebuild as a complete tree

Insert 2 &complete tree

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4 9

81 5

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6 91 4

AVL Trees 29

AVL - Good but not Perfect Balance

• AVL trees are height-balanced binary search trees

• Balance factor of a node– height(left subtree) - height(right subtree)

• An AVL tree has balance factor calculated at every node– For every node, heights of left and right subtree

can differ by no more than 1– Store current heights in each node

AVL Trees 30

Node Heights

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height of node = hbalance factor = hleft-hright

empty height = -1

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height=2 BF=1-0=1

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1 5

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Tree A (AVL) Tree B (AVL)

AVL Trees 31

Node Heights after Insert 7

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81 5

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height of node = hbalance factor = hleft-hright

empty height = -1

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balance factor 1-(-1) = 2

-1

Tree A (AVL) Tree B (not AVL)

AVL Trees 32

Insert and Rotation in AVL Trees

• Insert operation may cause balance factor to become 2 or –2 for some node – only nodes on the path from insertion point to

root node have possibly changed in height– So after the Insert, go back up to the root node by

node, updating heights– If a new balance factor (the difference hleft-hright) is

2 or –2, adjust tree by rotation around the node

AVL Trees 33

Single Rotation in an AVL Tree

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AVL Trees 34

Let the node that needs rebalancing be .

There are 4 cases: Outside Cases (require single rotation) : 1. Insertion into left subtree of left child of . 2. Insertion into right subtree of right child of . Inside Cases (require double rotation) : 3. Insertion into right subtree of left child of . 4. Insertion into left subtree of right child of .

The rebalancing is performed through four separate rotation algorithms.

Insertions in AVL Trees

AVL Trees 35

j

k

X Y

Z

Consider a validAVL subtree

AVL Insertion: Outside Case

h

hh

AVL Trees 36

j

k

XY

Z

Inserting into Xdestroys the AVL property at node j


h

h+1 h

AVL Trees 37

j

k

XY

Z

Do a “right rotation”


h

h+1 h

AVL Trees 38

j

k

XY

Z

Do a “right rotation”

Single right rotation

h

h+1 h

AVL Trees 39

jk

X Y Z

“Right rotation” done!(“Left rotation” is mirror symmetric)

Outside Case Completed

AVL property has been restored!

h

h+1

h

AVL Trees 40

j

k

X Y

Z

AVL Insertion: Inside Case Consider a validAVL subtree

h

hh

AVL Trees 41

Inserting into Y destroys theAVL propertyat node j

j

k

XY

Z

AVL Insertion: Inside Case

Does “right rotation”restore balance?

h

h+1h

AVL Trees 42

jk

X

YZ

“Right rotation”does not restorebalance… now k isout of balance


hh+1

h

AVL Trees 43

Consider the structureof subtree Y… j

k

XY

Z


h

h+1h

AVL Trees 44

j

k

XV

Z

W

i

Y = node i andsubtrees V and W


h

h+1h

h or h-1

AVL Trees 45

j

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Z

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i

AVL Insertion: Inside CaseWe will do a left-right “double rotation” . . .

AVL Trees 46

j

k

X V

ZW

i

Double rotation : first rotationleft rotation complete

AVL Trees 47

j

k

X V

ZW

i

Double rotation : second rotation

Now do a right rotation

AVL Trees 48

jk

X V ZW

i

Double rotation : second rotation

right rotation complete

Balance has been restored

hh h or h-1

AVL Trees 49

Implementation

balance (1,0,-1)

key

rightleft

No need to keep the height; just the difference in height, i.e. the balance factor; this has to be modified on the path of insertion even if you don’t perform rotations

Once you have performed a rotation (single or double) you won’t need to go back up the tree

AVL Trees 50

Example of Insertions in an AVL Tree

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Insert 5, 40

AVL Trees 51

Example of Insertions in an AVL Tree

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Now Insert 45

AVL Trees 52

Single rotation (outside case)

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Imbalance35 45

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Now Insert 34

AVL Trees 53

Double rotation (inside case)

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Imbalance

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Insertion of 34

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1 25 340

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Heap

• A heap is a binary tree T that stores a key-element pairs at its internal nodes.

• Properties :

• Min Heap: key(parent) < key(child)• Max Heap: key(parent) > key(child)]

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Example

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Example of a complete binary max-heap

• if B is a child node of A, then key(A) ≥ key(B).

• This implies that an element with the greatest key is always in the root node, and so such a heap is sometimes called a max-heap.

• Alternatively, if the comparison is reversed, the smallest element is always in the root node, which results in a min-heap.

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Max heap

• Terminate heap when– reach root– key parent is greater than key child

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6.8 and 6.10 no need

binary trees -2 chapter 6 1. 2 threaded trees (depth first) binary trees have a lot of wasted space:...

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