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Binary Search Tree Smt Genap

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**Outline Concept of Binary Search Tree (BST) BST operations**

Find Insert Remove Running time analysis of BST operations Smt Genap

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**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

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**Binary Search Tree: Examples**

7 9 2 1 5 6 3 Smt Genap

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**Binary Search Tree: Examples**

3 3 1 1 2 2 1 3 3 2 1 2 2 1 3 Smt Genap

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**Basic Operations FindMin, FindMax, Find Insert Remove**

Smt Genap

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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

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**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

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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

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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

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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

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**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

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**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

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

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**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

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Removing Leaf 8 12 4 6 1 3 5 Smt Genap

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**Removing Node With 1 Child**

8 12 4 6 1 3 5 Smt Genap

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**Removing Node With 1 Child**

8 12 4 6 1 3 5 Smt Genap

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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

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**Removing Node With 2 Children**

7 9 2 1 5 3 4 Smt Genap

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**Removing Node With 2 Children**

7 2 9 3 3 1 5 3 4 Smt Genap

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**Removing Node With 2 Children**

7 2 9 3 2 1 5 3 4 Smt Genap

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

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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

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**Find k-th element X X X SL SR SL SR SL SR k < SL + 1 k == SL + 1**

Smt Genap

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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

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**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

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**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

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