Examples

Transcription

1 Examples 1 Example [ proper binary tree properties ] Draw a proper binary tree with exactly 3 internal vertices and exactly 10 leaves. If not possible, explain why no such tree exists. Recall: the root counts as an internal vertex In a proper binary tree the following must be satisfied: E = I + 1 In the given case, I = 3 and E = 10. Clearly, these numbers do not satisfy the above requirement; hence, such a proper binary tree cannot be drown.

2 Examples (cont.) 2 Example [ proper binary tree properties ] Draw a proper binary tree with exactly 2 internal vertices and exactly 3 leaves. If not possible, explain why no such tree exists. Recall: the root counts as an internal vertex In a proper binary tree the following must be satisfied: E = I + 1 In the given case, I = 2 and E = 3, which satisfies the above requirement. The given tree would be:

3 Examples (cont.) 3 Example [ proper binary tree properties ] Draw a proper binary tree with height h = 3 and exactly 5 leaves. If not possible, explain why no such tree exists In a proper binary tree the following must be satisfied: In the given case, h = 3, E = 5, which satisfies the above requirement. Now, to draw the actual tree, we should first determine how many internal nodes the tree has: I = E-1 = 4. By examining different possible proper binary tress with 4 internal nodes, we conclude that the following tree satisfies h=3, and E=5 requirement. (Though, note, this is not the only such tree!!!) i i i i

4 Examples (cont.) 4 Example Let T be a full 3-ary tree with n 1 nodes (each parent has exactly children). Let H(n) denote the height of this tree. Prove: H(n) = (log 3 (2n+1)) - 1 Basic case: n=1 H(1) = (log 3 (2*1+1)) 1 = 0 OK! Case: n=2 - CANNOT BE!!! For every new n it must hold n = 3n prev + 1 every previous leaf becomes the root of a subtree, which turns out to be identical to the original tree 3x(size of old tree) + root

5 Binary Trees: Questions (cont.) 5 Case: n=4 H(4) = (log 3 (2 4+1)) 1 = 1 OK! Case: n H(n) = 1 + H(n prev ) = = 1+ log 3 (2n prev +1) 1 = = log 3 (2(n-1)/3+1) = = log 3 ((2n+1)/3) = = log 3 (2n+1)) - 1 OK!

6 Link Structure Implementation of Binary Tree 6 Node in Linked Structure object containing for Binary Trees 1) element 2) reference to parent 3) reference to the right child 4) reference to the left child if node is the root: reference to parent = null if node is external: references to children = null parent parent childrencontainer element left child element right child

7 Link Structure Implementation of Binary Tree (cont.) 7 Example [ binary tree and its linked list implementation ] A B C A B C D E D E

8 BTNode ADT: Implementation 8 BTNode Class generalization of Position ADT, i.e. implements Position interface public class BTNode<E> implements Position<E> { private E element; private Position<> parent, left, right; public BTNode(E element, Position<E> parent, Position<E> left, Position<E> right) { setelement(element); setparent(parent); setleft(left); setright(right); } public E getelement() { return element; } public void setelement(e e) { element = e; }

9 BTNode ADT: Implementation (cont.) 9 public Position<E> getleft() { return left; } public void setleft(position<e> v) { left = v; } public Position<E> getright() { return right; } public void setright(position<e> v) { right = v; } } public Position<E> getparent() { return parent; } public void setparent(position<e> v) { parent = v; } External Node: BTNode<E> extnode = new BTNode<E>(e, u, null, null) A null null null null null null

10 LinkedBinaryTree ADT: Implementation 10 LinkedBinaryTree Class implements BinaryTree interface plus the following 5 methods (see pp ) addroot addleft, addright remove attach public class LinkedBinaryTree<E> implements BinaryTree<E> { private BTPosition<E> root; /* reference to the root */ private int size; /* number of nodes */ public LinkedBinaryTree() { root = null; size = 0; } protected BTNode<E> validate(position<e> p) throws IllegalArgExc { if (!(p instanceof BTNode)) throw new IllegalArgExc( not valid ) BTNode<E> node = (BTNode<E>) p; if (node.getparent() == node) throw new IllegalArgExc( not in tree ); return node; }

11 LinkedBinaryTree ADT: Implementation (cont.) 11 public int size() { return size; } public Position<E> root() { return root; } public Position<E> parent(position<e> p) throw IllegalArgExc { Node<E> node = validate(p); return node.getparent(); } public Position<E> left(position<e> p) throw IllegalArgExc { Node<E> node = validate(p); return node.getleft(); } public Position<E> right(position<e> p) throw IllegalArgExc { } public Position<E> addroot (E e) throws IllegalStateException { if (!isempty()) throw new IllegalStateException( not empty... ) root = new BTNode<E>(e, null, null, null); size = 1; return root; }

12 LinkedBinaryTree ADT: Implementation (cont.) 12 /* Creates a new left child of Position p storing element e; returns its Position. */ private Position<E> addleft(position<e> p, E e) throws IllegalArgExc { BTnode<E> parent = validate(p); if (parent.getleft()!= null) throw new IllegalArgExc( has left ); BTNode<E> child = new BTNode<E>(e, parent, null, null); parent.setleft(child); size++; return child; } p e null null null null null

13 LinkedBinaryTree ADT: Implementation (cont.) 13 /* Removes nodes with zero or one child, returns node content. */ private E remove(position<e> p) throws IllegalArgExc { BTNode<E> node = validate(p); if (numchildren(p) == 2) throw new IllegalArgExc( has two children ); BTNode<E> child = (node.getleft()!= null? node.getleft() : node.getright() ); if (child!= null) child.setparent(node.getparent()); if (node == root) root = child; // child becomes root else { BTNode<E> parent = node.getparent(); if (node == parent.getleft()) parent.setleft(child); else parent.setright(child); } w null null v null size--; E temp = node.getelement(); v node.setelement(null); node.setparent(node); w null null null return temp; } null null

14 LinkedBinaryTree ADT: Implementation (cont.) 14 /* Attaches trees t1 and t2 to be subtrees of an external node p. */ private void attach(position<e> p, LinkedBinaryTree<E> t1, LinkedBinaryTree<E> t2) throws InvalidArgExc { BTNode<E> node = validate(p); if (isinternal(p)) throw new IllegalArgExc( must be a leaf ); size += t1.size() + t2.size(); if (!t1.isempty() { t1.root.setparent(node); node.setleft(t1.root); t1.root = null; t1.size = 0; } v /* similar step for T2 */ null null null null } T1 T2

15 LinkedBinaryTree ADT: Implementation (cont.) 15 Run Times Good! all methods, except positions and elements run in constant O(1) time Space Complexity Good! only O(n), since there is one BTNode object per every node of the tree no empty slots as in array-based implementation Method iterator, positions size, isempty replace root, left, right, parent, children, sibling hasleft, hasright, isinternal, isexternal, isroot addroot, addleft, addright, attach, remove Time O(n) O(1) O(1) O(1) O(1) O(1)

16 LinkedBinaryTree ADT: Implementation (cont.) 16 Example 3 [ creating a tree ] LinkedBinaryTree<String> t = new LinkedBinaryTree<String>(); t.addroot( Albert ); t.addleft(t.root(), Betty ); t.addright(t.root(), Chris ); t.addleft(t.root().left(), David ); t.addright(t.root().left(), Elvis ); Albert t Betty Chris David Elvis