# Binary Search Tree Smt Genap 2011-2012.

## Presentation on theme: "Binary Search Tree Smt Genap 2011-2012."— Presentation transcript:

Binary Search Tree Smt Genap

Outline Concept of Binary Search Tree (BST) BST operations
Find Insert Remove Running time analysis of BST operations Smt Genap

Binary Search Tree: Properties
Elements have keys (no duplicates allowed). 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. The keys must be comparable. X <X >X Smt Genap

Binary Search Tree: Examples
7 9 2 1 5 6 3 Smt Genap

Binary Search Tree: Examples
3 3 1 1 2 2 1 3 3 2 1 2 2 1 3 Smt Genap

Basic Operations FindMin, FindMax, Find Insert Remove
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FindMin Find node with the smallest value Algorithm: Code:
Keep going left until you reach a dead end! Code: BinaryNode<Type> findMin(BinaryNode<Type> t) { if (t != null) while (t.left != null) t = t.left; return t; } Smt Genap

FindMax Find node with the largest value Algorithm: Code:
Keep going right until you reach a dead end! Code: BinaryNode<Type> findMax(BinaryNode<Type> t) { if (t != null) while (t.right != null) t = t.right; return t; } Smt Genap

Find You are given an element to find in a BST. If it exists, return the node. If not, return null. Algorithm? Code? 7 9 2 1 5 6 3 Smt Genap

Find: Implementation BinaryNode<Type> find(Type x, BinaryNode<T> t) { while(t!=null) if(x.compareTo(t.element)<0) t = t.left; else if(x.compareTo(t.element)>0) t = t.right; else return t; // Match } return null; // Not found Smt Genap

Insertion: Principle When inserting a new element into a binary search tree, it will always become a leaf node. 10 2 3 15 1 5 6 12 14 Smt Genap

Insertion: Algorithm To insert X into a binary search tree:
Start from the root If the value of X < the value of the root: X should be inserted in the left sub-tree. If the value of X > the value of the root: X should be inserted in the right sub-tree. Remember that a sub-tree is also a tree. We can implement this recursively! Smt Genap

Insertion: Implementation
BinaryNode<Type> insert(Type x, BinaryNode<Type> t) { if (t == null) t = new BinaryNode<Type>(x); else if(x.compareTo(t.element)<0) t.left = insert (x, t.left); else if(x.compareTo(t.element)>0) t.right = insert (x, t.right); else throw new DuplicateItemException(x); return t; } Smt Genap

Removing An Element 8 4 5 12 1 6 3 4 6 5 Smt Genap

Removing An Element: Algorithm
If the node is a leaf, simply delete it. If the node has one child, adjust parent’s child reference to bypass the node. If the node has two children: Replace the node’s element with the smallest element in the right subtree and then remove that node, or Replace the node’s element with the largest element in the left subtree and then remove that node Introduces new sub-problems: removeMin: Alternatively, removeMax Smt Genap

Removing Leaf 8 12 4 6 1 3 5 Smt Genap

Removing Node With 1 Child
8 12 4 6 1 3 5 Smt Genap

Removing Node With 1 Child
8 12 4 6 1 3 5 Smt Genap

removeMin BinaryNode<Type> removeMin(BinaryNode<Type> t) { if (t == null) throw new ItemNotFoundException(); else if (t.left != null) t.left = removeMin(t.left); return t; } else return t.right; Smt Genap

Removing Node With 2 Children
7 9 2 1 5 3 4 Smt Genap

Removing Node With 2 Children
7 2 9 3 3 1 5 3 4 Smt Genap

Removing Node With 2 Children
7 2 9 3 2 1 5 3 4 Smt Genap

Removing Root 7 2 3 12 1 5 4 10 14 9 11 9 Smt Genap

Remove BinaryNode<Type> remove(Type x, BinaryNode<Type> t) { if (t == null) throw new ItemNotFoundException(); if (x.compareTo(t.element)<0) t.left = remove(x, t.left); else if(x.compareTo(t.element)>0) 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 if(t.left!=null) t=t.left; else t=t.right; return t; Smt Genap

Find k-th element X X X SL SR SL SR SL SR k < SL + 1 k == SL + 1
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Find k-th element BinaryNode<Type> findKth(int k, BinaryNode<Type> 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); } Smt Genap

Analysis Running time for: Average case: O(log n) Worst case: O(n)
Insert? Find min? Remove? Find? Average case: O(log n) Worst case: O(n) Smt Genap

Summary Binary Search Tree maintains the order of the tree.
Each node should be comparable All operations take O(log n) - average case, when the tree is equally balanced. All operations will take O(n) - worst case, when the height of the tree equals the number of nodes. Smt Genap