Binary Search Tree Smt Genap 2011-2012

Outline Concept of Binary Search Tree (BST) BST operations Find Insert Remove Running time analysis of BST operations Smt Genap 2011-2012

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 2011-2012

Binary Search Tree: Examples 7 9 2 1 5 6 3 Smt Genap 2011-2012

Binary Search Tree: Examples 3 3 1 1 2 2 1 3 3 2 1 2 2 1 3 Smt Genap 2011-2012

Basic Operations FindMin, FindMax, Find Insert Remove Smt Genap 2011-2012

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 2011-2012

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 2011-2012

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 2011-2012

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 2011-2012

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 2011-2012

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 2011-2012

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 2011-2012

Removing An Element 8 4 5 12 1 6 3 4 6 5 Smt Genap 2011-2012

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 2011-2012

Removing Leaf 8 12 4 6 1 3 5 Smt Genap 2011-2012

Removing Node With 1 Child 8 12 4 6 1 3 5 Smt Genap 2011-2012

Removing Node With 1 Child 8 12 4 6 1 3 5 Smt Genap 2011-2012

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 2011-2012

Removing Node With 2 Children 7 9 2 1 5 3 4 Smt Genap 2011-2012

Removing Node With 2 Children 7 2 9 3 3 1 5 3 4 Smt Genap 2011-2012

Removing Node With 2 Children 7 2 9 3 2 1 5 3 4 Smt Genap 2011-2012

Removing Root 7 2 3 12 1 5 4 10 14 9 11 9 Smt Genap 2011-2012

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 2011-2012

Find k-th element X X X SL SR SL SR SL SR k < SL + 1 k == SL + 1 Smt Genap 2011-2012

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 2011-2012

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 2011-2012

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 2011-2012