Objectives Memahami sifat dari Binary Search Tree (BST) Struktur - - PowerPoint PPT Presentation

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Oct 25, 2023 279 likes •378 views

Objectives Memahami sifat dari Binary Search Tree (BST) Struktur Data & Algoritme Memahami operasi-operasi pada BST Memahami kelebihan dan kekurangan dari BST ( Data Structures & Algorithms ) Binary Search Tree Denny (

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1 Struktur Data & Algoritme ( Data Structures & Algorithms)

Denny (denny@cs.ui.ac.id) Suryana Setiawan (setiawan@cs.ui.ac.id)

Fakultas I lm u Kom puter Universitas I ndonesia Sem ester Genap - 2 0 0 4 / 2 0 0 5

Version 2 .0 - I nternal Use Only

Binary Search Tree

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Objectives

Memahami sifat dari Binary Search Tree (BST) Memahami operasi-operasi pada BST Memahami kelebihan dan kekurangan dari BST

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Outline

Properties of Binary Search Tree (BST) Operation Insert find remove

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Properties of Binary Search Tree

For every node X in the tree, the values of all the keys

in the left subtree are smaller than the key in X and the values of all the keys in the right subtree are larger than the key in X.

So, the key should be comparable.

X <X >X

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Binary Search Tree

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Binary Search Tree

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Basic Operations

insert findMin and findMax remove

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Insertion

Penyisipan sebuah elemen baru dalam binary search

tree, elemen tersebut pasti akan menjadi leaf

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Insertion: algorithm

Insert X into a binary search tree: start from the root. If the value of X is less than the value of the root, then

X should be inserted on the left sub-tree.

On the other hand, if the value of X is greater than the

value of the root, then X should be inserted on the right sub-tree.

Remember that, a sub tree is also a tree. So, the

problem to insert an element in the sub-tree is same as the problem to insert an element in the root.

So? We can attack this problem with recursive approach.

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Insertion

BinaryNode insert(int x, BinaryNode t) { if (t == null) { t = new BinaryNode (x, null, null); } else if (x < t.element) { t.left = insert (x, t.left); } else if (x > t.element) { t.right = insert (x, t.right); } else { throw new DuplicateItem(“exception”); } return t; }

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FindMin

Mencari node yang memiliki nilai terkecil. Algorithm: ke kiri terus sampai buntu….:) Code:

BinaryNode findMin (BinaryNode t) { if (t == null) throw exception; while (t.left != null) { t = t.left; } return t; }

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FindMax

Mencari node yang memiliki nilai terbesar Algorithm? Code?

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Find

Diberikan sebuah nilai yang harus dicari dalam

sebuah BST. Jika ada elemen tersebut, return node

tersebut. Jika tidak ada, return null.

Algorithm? Code?

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Remove

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Remove

if the node is a leaf (has no child), no problemo…

delete it immediately

if the node has one child: its parent adjusts a child

reference to bypass the node.

if the node has two children? replace the item in this node with the smallest item in

the right subtree and then remove that node, or

replace the item in this node with the biggest item in

the left subtree and then remove that node

introduce new sub-problems: removeMin, removeMax

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Removing 6

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3

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After 6 removed

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3

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removeMin

BinaryNode removeMin(BinaryNode t) { if (t == null) throw exception; if (t.left != null) { t.left = removeMin (t.left); } else { t = t.right; } return t; }

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removeMax

code?

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Removing 2

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6

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After 2 deleted

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3 X

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Removing Root

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Remove

BinaryNode remove(int x, BinaryNode t) { if (t == null) throw exception; if (x < t.element) { t.left = remove(x, t.left); } else if (x > t.element) { t.right = remove(x, t.right); } else if (t.left != null && t.right != null) { t.element = findMin(t.right).element; t.right = removeMin(t.right); } else { t = (t.left != null) ? t.left : t.right; } return t; }

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SL SR X k < SL + 1 SL SR X k == SL + 1 SL SR X k > SL + 1

Find k-th element

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Find k-th element

BinaryNode findKth(int k, BinaryNode t) { if (t == null) throw exception; int leftSize = (t.left != null) ? t.left.size : 0; if (k <= leftSize ) { return findKth (k, t.left); } else if (k == leftSize + 1) { return t; } else { return findKth ( k - leftSize - 1, t.right); } }

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Analysis

Runnning time for: insert? Find min? remove? Find? Worst case: O(n)

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