Upload
angelica-finkle
View
222
Download
1
Tags:
Embed Size (px)
Citation preview
AVL Trees 2
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
An example of an AVL tree where the heights are shown next to the nodes:
AVL Trees 3
Balanced nodes • A internal node is balanced if the heights of its
two children differ by at most 1. • Otherwise, such an internal node is
unbalanced.
AVL Trees 4
Height of an AVL Tree• Fact: The height of an AVL tree storing n keys is O(log n).• Proof: Let us bound n(h): the minimum number of internal
nodes of an AVL tree of height h.• We easily see that n(1) = 1 and n(2) = 2• For n > 2, an AVL tree of height h contains the root node, one
AVL subtree of height n-1 and another of height n-2.• That is, n(h) = 1 + n(h-1) + n(h-2)• Knowing n(h-1) > n(h-2), we get n(h) > 2n(h-2). So
n(h) > 2n(h-2), n(h) > 4n(h-4), n(h) > 8n(n-6), … (by induction),n(h) > 2in(h-2i)>2 {h/2 -1} (1) = 2 {h/2 -1}
• Solving the base case we get: n(h) > 2 h/2-1
• Taking logarithms: h < 2log n(h) +2• Thus the height of an AVL tree is O(log n)
3
4 n(1)
n(2)
h-1h-2
AVL Trees 5
Insertion in an AVL Tree• Insertion is as in a binary search tree• Always done by expanding an external node.• Example:
44
17 78
32 50 88
48 62
54w
b=x
a=y
c=z
44
17 78
32 50 88
48 62
before insertion after insertionIt is no longer balanced
AVL Trees 6
Names of important nodes• w: the newly inserted node. (insertion process follow the binary search
tree method)• The heights of some nodes in T might be increased after inserting a node.
– Those nodes must be on the path from w to the root. – Other nodes are not effected.
• z: the first node we encounter in going up from w toward the root such that z is unbalanced.
• y: the child of z with higher height.– y must be an ancestor of w. (why? Because z in unbalanced after inserting w)
• x: the child of y with higher height.– The height of the sibling of x is smaller than that of x. (Otherwise, the height of
y cannot be increased.)– x must be an ancestor of w.
See the figure in the last slide.
AVL Trees 7
Algorithm restructure(x): Input: A node x of a binary search tree T that has both
parent y and grand-parent z.Output: Tree T after a trinode restructuring.1. Let (a, b, c) be the list (increasing order) of nodes x, y,
and z. Let T0, T1, T2 T3 be a left-to-right (inorder) listing of the four subtrees of x, y, and z not rooted at x, y, or z.
2. Replace the subtree rooted at z with a new subtree rooted at b..
3. Let a be the left child of b and let T0 and T1 be the left and right subtrees of a, respectively.
4. Let c be the right child of b and let T2 and T3 be the left and right subtrees of c, respectively.
AVL Trees 8
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
T3T2T1
T0
a = xb = y
c = zsingle rotation
AVL Trees 9
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
T3T1
T2
T0 T3
T1T0 T2
c = zb = x
a = y
AVL Trees 10
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
AVL Trees 11
Theorem: –One restructure operation is enough to
ensure that the whole tree is balanced.–Proof: Left to the readers.
Quiz 3
Misalkan diberikan data sebagai berikut :(1,”A”), (3,”C”),(2,”B”),E,I,D,F,G,H, J1. Susun data di atas ke dalam CBT berdasarkan indeks
menggunakan array. 2. Susun data diatas sebagai representasi PQ menggunakan
MinHeapTree berdasarkan Kunci3. Hapus step by step dari HeapTree4. Susun data di atas ke dalam AVL tree
AVL Trees 12
• blc = tki – tka• t = max(tki, tka)
AVL Trees 13
Dt pKi pKa tKi tKa Blc t
NodeAVL
Object dtNodeAVL pKi, pKaint tKi, tkA, blc, tNodeAVL(Object dt)int getBalance()
public void sisipData(int dt){size++; // tambah jumlah nodeNodeAVL baru = new NodeAVL(dt); // buat node baru
if (root == null) root = baru; // jika root masih nullelse{ // jika root tidak null
NodeAVL pKini, pInduk = null;pKini = root;StackAVL st = new StackAVL(size); // buat stack
// telusuri node untuk membentuk path dari root ke posisiwhile(pKini != null){
st.push(pKini); // simpan node pKini di stackpInduk = pKini; if (dt < pKini.data) pKini = pKini.anakKiri;else pKini = pKini.anakKanan;
}
AVL Trees 15
//sisipkan nodeif (!isNotSeimbang(pInduk)){ // jika sudah seimbang
if(dt < pInduk.data) {pInduk.anakKiri = baru;
}else {
pInduk.anakKanan = baru;}
///// -----}
}else { // jika belum seimbang --> seimbangkan dulu
seimbangkan(pInduk);}
}}
AVL Trees 16
//update date node induk-induknyaNodeAVL pathNode;while(!st.isEmpty()) {
pathNode = st.pop();
if(pathNode.anakKiri == null) pathNode.tKiri = 0;else pathNode.tKiri = pathNode.anakKiri.tinggi;
if(pathNode.anakKanan == null) pathNode.tKanan = 0;
else pathNode.tKanan = pathNode.anakKanan.tinggi;
pathNode.seimbang = Math.abs(pathNode.tKanan - pathNode.tKiri);
pathNode.tinggi = 1 + max(pathNode.tKiri, pathNode.tKanan);
if (isNotSeimbang(pathNode)){System.out.println("TESSSS");seimbangkan(pathNode);
}
AVL Trees 17
AVL Trees 18
Removal in an AVL Tree• Removal 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.
• Example: 44
17
7832 50
8848
62
54
44
17
7850
8848
62
54
before deletion of 32 after deletion
AVL Trees 19
Rebalancing after a Removal
• Let z be the first unbalanced node encountered while travelling up the tree from w. Also,
• let y be the child of z with the larger height,• let x be the child of y defined as follows;
– If one of the children of y is taller than the other, choose x as the taller child of y.
– If both children of y have the same height, select x be the child of y on the same side as y (i.e., if y is the left child of z, then x is the left child of y; and if y is the right child of z then x is the right child of y.)
AVL Trees 20
Rebalancing after a Removal• We perform restructure(x) to restore balance at z.• As this restructuring may upset the balance of another
node higher in the tree, we must continue checking for balance until the root of T is reached
44
17
7850
8848
62
54
w
c=x
b=y
a=z
44
17
78
50 88
48
62
54
AVL Trees 21
Unbalanced after restructuring
44
17
7850
88
62w
c=x
b=y
a=z
44
17
78
50 88
62
32
1 1
h=5h=5h=3h=4
Unbalanced balanced
AVL Trees 22
Rebalancing after a Removal• We perform restructure(x) to restore balance at z.• As this restructuring may upset the balance of another
node higher in the tree, we must continue checking for balance until the root of T is reached
44
17
7850
8848
62
54
w
c=x
b=y
a=z
44
17
78
50 88
48
62
54
AVL Trees 23
Running Times for AVL Trees
• a single restructure is O(1)– using a linked-structure binary tree
• find is O(log n)– height of tree is O(log n), no restructures needed
• insert is O(log n)– initial find is O(log n)– Restructuring up the tree, maintaining heights is O(log n)
• remove is O(log n)– initial find is O(log n)– Restructuring up the tree, maintaining heights is O(log n)
AVL Trees 25
Divide-and-Conquer (§ 10.1.1)
• Divide-and conquer is a general algorithm design paradigm:– Divide: divide the input data S
in two disjoint subsets S1 and S2
– Recur: solve the subproblems associated with S1 and S2
– Conquer: combine the solutions for S1 and S2 into a solution for S
• The base case for the recursion are subproblems of size 0 or 1
• Merge-sort is a sorting algorithm based on the divide-and-conquer paradigm
• Like heap-sort– It uses a comparator– It has O(n log n) running
time• Unlike heap-sort
– It does not use an auxiliary priority queue
– It accesses data in a sequential manner (suitable to sort data on a disk)
AVL Trees 26
Merge-Sort (§ 10.1)
• Merge-sort on an input sequence S with n elements consists of three steps:– Divide: partition S into two
sequences S1 and S2 of about n/2 elements each
– Recur: recursively sort S1 and S2
– Conquer: merge S1 and S2
into a unique sorted sequence
Algorithm mergeSort(S, C)Input sequence S with n
elements, comparator C Output sequence S sorted
according to Cif S.size() > 1
(S1, S2) partition(S, n/2)
mergeSort(S1, C)
mergeSort(S2, C)
S merge(S1, S2)
AVL Trees 27
Merging Two Sorted Sequences• The conquer step of
merge-sort consists of merging two sorted sequences A and B into a sorted sequence S containing the union of the elements of A and B
• Merging two sorted sequences, each with n/2 elements and implemented by means of a doubly linked list, takes O(n) time
Algorithm merge(A, B)Input sequences A and B with
n/2 elements each
Output sorted sequence of A B
S empty sequence
while A.isEmpty() B.isEmpty()if A.first().element() <
B.first().element()
S.insertLast(A.remove(A.first()))else
S.insertLast(B.remove(B.first()))while A.isEmpty()S.insertLast(A.remove(A.first()))while B.isEmpty()S.insertLast(B.remove(B.first()))return S
AVL Trees 28
Merge-Sort Tree• An execution of merge-sort is depicted by a binary tree
– each node represents a recursive call of merge-sort and stores• unsorted sequence before the execution and its partition• sorted sequence at the end of the execution
– the root is the initial call – the leaves are calls on subsequences of size 0 or 1
7 2 9 4 2 4 7 9
7 2 2 7 9 4 4 9
7 7 2 2 9 9 4 4
AVL Trees 29
Execution Example
• Partition
7 2 9 4 2 4 7 9 3 8 6 1 1 3 8 6
7 2 2 7 9 4 4 9 3 8 3 8 6 1 1 6
7 7 2 2 9 9 4 4 3 3 8 8 6 6 1 1
7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9
AVL Trees 30
Execution Example (cont.)
• Recursive call, partition
7 2 9 4 2 4 7 9 3 8 6 1 1 3 8 6
7 2 2 7 9 4 4 9 3 8 3 8 6 1 1 6
7 7 2 2 9 9 4 4 3 3 8 8 6 6 1 1
7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9
AVL Trees 31
Execution Example (cont.)
• Recursive call, partition
7 2 9 4 2 4 7 9 3 8 6 1 1 3 8 6
7 2 2 7 9 4 4 9 3 8 3 8 6 1 1 6
7 7 2 2 9 9 4 4 3 3 8 8 6 6 1 1
7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9
AVL Trees 32
Execution Example (cont.)
• Recursive call, base case
7 2 9 4 2 4 7 9 3 8 6 1 1 3 8 6
7 2 2 7 9 4 4 9 3 8 3 8 6 1 1 6
7 7 2 2 9 9 4 4 3 3 8 8 6 6 1 1
7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9
AVL Trees 33
Execution Example (cont.)
• Recursive call, base case
7 2 9 4 2 4 7 9 3 8 6 1 1 3 8 6
7 2 2 7 9 4 4 9 3 8 3 8 6 1 1 6
7 7 2 2 9 9 4 4 3 3 8 8 6 6 1 1
7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9
AVL Trees 34
Execution Example (cont.)
• Merge
7 2 9 4 2 4 7 9 3 8 6 1 1 3 8 6
7 2 2 7 9 4 4 9 3 8 3 8 6 1 1 6
7 7 2 2 9 9 4 4 3 3 8 8 6 6 1 1
7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9
AVL Trees 35
Execution Example (cont.)
• Recursive call, …, base case, merge
7 2 9 4 2 4 7 9 3 8 6 1 1 3 8 6
7 2 2 7 9 4 4 9 3 8 3 8 6 1 1 6
7 7 2 2 3 3 8 8 6 6 1 1
7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9
9 9 4 4
AVL Trees 36
Execution Example (cont.)
• Merge
7 2 9 4 2 4 7 9 3 8 6 1 1 3 8 6
7 2 2 7 9 4 4 9 3 8 3 8 6 1 1 6
7 7 2 2 9 9 4 4 3 3 8 8 6 6 1 1
7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9
AVL Trees 37
Execution Example (cont.)
• Recursive call, …, merge, merge
7 2 9 4 2 4 7 9 3 8 6 1 1 3 6 8
7 2 2 7 9 4 4 9 3 8 3 8 6 1 1 6
7 7 2 2 9 9 4 4 3 3 8 8 6 6 1 1
7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9
AVL Trees 38
Execution Example (cont.)
• Merge
7 2 9 4 2 4 7 9 3 8 6 1 1 3 6 8
7 2 2 7 9 4 4 9 3 8 3 8 6 1 1 6
7 7 2 2 9 9 4 4 3 3 8 8 6 6 1 1
7 2 9 4 3 8 6 1 1 2 3 4 6 7 8 9
AVL Trees 39
Analysis of Merge-Sort• The height h of the merge-sort tree is O(log n)
– at each recursive call we divide in half the sequence, • The overall amount or work done at the nodes of depth i is O(n)
– we partition and merge 2i sequences of size n/2i – we make 2i+1 recursive calls
• Thus, the total running time of merge-sort is O(n log n)
depth #seqs
size
0 1 n
1 2 n/2
i 2i n/2i
… … …
AVL Trees 40
Summary of Sorting Algorithms
Algorithm Time Notes
selection-sort
O(n2) slow in-place for small data sets (< 1K)
insertion-sort
O(n2) slow in-place for small data sets (< 1K)
heap-sort O(n log n)
fast in-place for large data sets (1K —
1M)
merge-sort O(n log n) fast sequential data access for huge data sets (> 1M)