© 2004 goodrich, tamassia binary search trees1 csc 212 lecture 18: binary and avl trees

Binary Search Trees 1© 2004 Goodrich, Tamassia© 2004 Goodrich, Tamassia

CSC 212

Lecture 18: Binary and AVL Trees


Ordered DictionariesKeys are assumed to come from a total order.New operations: first(): first entry in the dictionary ordering last(): last entry in the dictionary ordering successor(k): entry with smallest key

greater than or equal to k; increasing order predecessor(k): entry with largest key less

than or equal to k; decreasing order


Binary Search (§ 8.3.3)Binary search can perform operation find(k) on a dictionary implemented by means of an array-based sequence, sorted by key

similar to the high-low game at each step, the number of candidate items is halved terminates after O(log n) steps

Example: find(7)

1 3 4 5 7 8 9 11 14 16 18 19

1 3 4 5 7 8 9 11 14 16 18 19

1 3 4 5 7 8 9 11 14 16 18 19

1 3 4 5 7 8 9 11 14 16 18 19

0

0

0

0

ml h

ml h

ml h

lm h


Search Tables

A search table is a dictionary implemented by means of a sorted sequence

Store the items of the dictionary in an array-based sequence, sorted by key

Performance: find takes O(log n) time, using binary search insert takes O(n) time since in the worst case we have

to shift n2 items to make room for the new item remove take O(n) time since in the worst case we have

to shift n2 items to compact the items after the removal

Lookup table effective only for dictionaries of small size or for dictionaries on which searches are the most common operations


Binary Search Trees (§ 9.1)A binary search tree is a binary tree storing keys (or key-value entries) at its internal nodes and satisfying the following property:

Let u, v, and w be three nodes such that u is in the left subtree of v and w is in the right subtree of v. We have key(u) key(v) key(w)

External nodes do not store items

An inorder traversal of a binary search trees visits the keys in increasing order

6

92

41 8


Search (§ 9.1.1)To search for a key k, we trace a downward path starting at the rootThe next node visited depends on the outcome of the comparison of k with the key of the current nodeIf we reach a leaf, the key is not found and we return nullExample: find(4):

Call TreeSearch(4,root)

Algorithm TreeSearch(k, v)if T.isExternal (v)

/* throw exception */if k key(v)

return TreeSearch(k, T.left(v))else if k key(v)

return velse { k key(v) }

return TreeSearch(k, T.right(v))

6

92

41 8


Search (§ 9.1.1)Example: find(5):

Call TreeSearch(5,root)

Algorithm TreeSearch(k, v)if T.isExternal (v)

/* throw exception */if k key(v)

return TreeSearch(k, T.left(v))else if k key(v)

return velse { k key(v) }

return TreeSearch(k, T.right(v))

6

92

41 8


InsertionTo perform operation insert(k, o), we search for key k (using TreeSearch)Assume k is not already in the tree, and let let w be the leaf reached by the searchWe insert k at node w and expand w into an internal nodeExample: insert(5, o)

6

92

41 8

6

92

41 8

5

w

w


DeletionTo perform operation remove(k), we search for key kAssume key k is in the tree, and let let v be the node storing kIf node v has a leaf child w, we remove v and w from the tree with operation removeExternal(w), which removes w and its parentExample: remove(4)

6

92

41 8

5

vw

6

92

51 8


Deletion (cont.)We consider the case where the key k to be removed is stored at a node v whose children are both internal

we find the internal node w that follows v in an inorder traversal

we copy key(w) into node v we remove node w and its

left child z (which must be a leaf) by means of operation removeExternal(z)

Example: remove(3)

3

1

8

6 9

5

v

w

z

2

5

1

8

6 9

v

2


PerformanceConsider a dictionary with n items implemented by means of a binary search tree of height h

the space used is O(n) methods find, insert

and remove take O(h) time

The height h is O(n) in the worst case and O(log n) in the best case


Your Turn

Insert 4, 2, 6, 1, 3, 5, 7 into a treeThen remove nodes in this order:4, 2, 6, 1, 3, 5, 7


AVL Tree Definition (§ 9.2)

AVL trees are balanced.An AVL Tree is a binary search tree such that for every internal node v of T, the heights of the children of v can differ by at most 1.

88

44

17 78

32 50

48 62

2

4

1

1

2

3

1

1

Example AVL tree where the heights are shown next to the nodes


Insertion in an AVL TreeInsertion is as in a binary search treeAlways done by expanding an external node.Consider insert(54, 0):

44

17 78

32 50 88

48 62

54

44

17 78

32 50 88

48 62

before insertion after insertion


Trinode Restructuringlet (a,b,c) be an inorder listing of x, y, zperform the rotations needed to make b the topmost node of the three

b=y

a=z

c=x

T0

T1

T2 T3

b=y

a=z c=x

T0 T1 T2 T3

c=y

b=x

a=z

T0

T1 T2

T3b=x

c=ya=z

T0 T1 T2 T3

case 1: single rotation(a left rotation about a)

case 2: double rotation(a right rotation about c, then a left rotation about a)

(other two cases are symmetrical)


Insertion Example, continued

88

44

17 78

32 50

48 62

2

5

1

1

3

4

2

1

54

1

T0T2

T3

x

y

z

2

3

4

5

67

1

88

44

17

7832 50

48

622

4

1

1

2 2

3

154

1

T0 T1

T2

T3

x

y z

unbalanced...

...balanced

1

2

3

4

5

6

7

T1


Restructuring (as Single Rotations)

Single Rotations:

T0T1

T2

T3

c = xb = y

a = z

T0 T1 T2

T3

c = xb = y

a = zsingle rotation

T3T2

T1

T0

a = xb = y

c = z

T0T1T2

T3

a = xb = y

c = zsingle rotation


Restructuring (as Double Rotations)

double rotations:

double rotationa = z

b = xc = y

T0T2

T1

T3 T0

T2T3T1

a = zb = x

c = y

double rotationc = z

b = xa = y

T0T2

T1

T3 T0

T2T3 T1

c = zb = x

a = y


Removal in an AVL TreeRemoval begins as in a binary search tree, which means the node removed will become an empty external node. Its parent, w, may cause an imbalance.Remove(32):

44

17

7832 50

8848

62

54

44

17

7850

8848

62

54

before deletion of 32 after deletion


Rebalancing after a Removal

Let z be the first unbalanced node encountered while travelling up tree after removing a node. Let y be the child of z with the larger height, and x be the child of y with the larger height.restructure(x) restores balance at z.Restructuring may unbalance node higher in the tree, so must continue checking for balance (and restructuring where needed) until root is reached44

17

7850

8848

62

54

w

c=x

b=y

a=z

44

17

78

50 88

48

62

54


Running Times for AVL Trees

Single restructure is O(1) Work is constant for height of tree

find is O(log n) height of tree is O(log n)

insert is _____________ initial find is O(log n) Restructuring up the tree – could do at most

_______________________ workremove is _______________ initial find is O(log n) Restructuring up the tree – could do at most

_______________________ work


Your Turn

Insert 1, 2, 3, 4, 5, 6, 7, 8 into an AVL treeThen remove 1, 2, 3, 4, 5, 6, 7, 8 into AVL tree


Daily Quiz

I have a “friend” who claims that the order nodes are entered into an AVL tree does not matter, the end tree is always the same Provide a small example proving my

“friend” wrong

© 2004 goodrich, tamassia binary search trees1 csc 212 lecture 18: binary and avl trees

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