Tree Traversal Techniques; Heaps.ppt

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1、CS 103,1,Tree Traversal Techniques; Heaps,Tree Traversal Concept Tree Traversal Techniques: Preorder, Inorder, Postorder Full Trees Almost Complete Trees Heaps,CS 103,2,Binary-Tree Related Definitions,The children of any node in a binary tree are ordered into a left child and a right child A node ca

2、n have a left anda right child, a left childonly, a right child only,or no children The tree made up of a leftchild (of a node x) and all itsdescendents is called the left subtree of x Right subtrees are defined similarly,10,CS 103,3,A Binary-tree Node Class,class TreeNode public: typedef int dataty

3、pe; TreeNode(datatype x=0, TreeNode *left=NULL, TreeNode *right=NULL)data=x; this-left=left; this-right=right; ;datatype getData( ) return data;TreeNode *getLeft( ) return left;TreeNode *getRight( ) return right;void setData(datatype x) data=x;void setLeft(TreeNode *ptr) left=ptr;void setRight(TreeN

4、ode *ptr) right=ptr; private:datatype data; / different data type for other appsTreeNode *left; / the pointer to left childTreeNode *right; / the pointer to right child ;,CS 103,4,Binary Tree Class,class Tree public: typedef int datatype; Tree(TreeNode *rootPtr=NULL)this-rootPtr=rootPtr;TreeNode *se

5、arch(datatype x);bool insert(datatype x); TreeNode * remove(datatype x);TreeNode *getRoot()return rootPtr;Tree *getLeftSubtree(); Tree *getRightSubtree();bool isEmpty()return rootPtr = NULL;private: TreeNode *rootPtr; ;,CS 103,5,Binary Tree Traversal,Traversal is the process of visiting every node o

6、nce Visiting a node entails doing some processing at that node, but when describing a traversal strategy, we need not concern ourselves with what that processing is,CS 103,6,Binary Tree Traversal Techniques,Three recursive techniques for binary tree traversal In each technique, the left subtree is t

7、raversed recursively, the right subtree is traversed recursively, and the root is visited What distinguishes the techniques from one another is the order of those 3 tasks,CS 103,7,Preoder, Inorder, Postorder,In Preorder, the rootis visited before (pre)the subtrees traversals In Inorder, the root isv

8、isited in-between left and right subtree traversal In Preorder, the rootis visited after (pre)the subtrees traversals,Preorder Traversal: Visit the root Traverse left subtree Traverse right subtree,Inorder Traversal: Traverse left subtree Visit the root Traverse right subtree,Postorder Traversal: Tr

9、averse left subtree Traverse right subtree Visit the root,CS 103,8,Illustrations for Traversals,Assume: visiting a nodeis printing its label Preorder: 1 3 5 4 6 7 8 9 10 11 12 Inorder:4 5 6 3 1 8 7 9 11 10 12 Postorder:4 6 5 3 8 11 12 10 9 7 1,CS 103,9,Illustrations for Traversals (Contd.),Assume: v

10、isiting a nodeis printing its data Preorder: 15 8 2 6 3 711 10 12 14 20 27 22 30 Inorder: 2 3 6 7 8 10 1112 14 15 20 22 27 30 Postorder: 3 7 6 2 10 1412 11 8 22 30 27 20 15,CS 103,10,Code for the Traversal Techniques,The code for visitis up to you toprovide, dependingon the application A typical exa

11、mplefor visit() is toprint out the data part of its inputnode,void inOrder(Tree *tree)if (tree-isEmpty( ) return;inOrder(tree-getLeftSubtree( );visit(tree-getRoot( );inOrder(tree-getRightSubtree( ); ,void preOrder(Tree *tree)if (tree-isEmpty( ) return;visit(tree-getRoot( );preOrder(tree-getLeftSubtr

12、ee();preOrder(tree-getRightSubtree(); ,void postOrder(Tree *tree)if (tree-isEmpty( ) return;postOrder(tree-getLeftSubtree( );postOrder(tree-getRightSubtree( );visit(tree-getRoot( ); ,CS 103,11,Application of Traversal Sorting a BST,Observe the output of the inorder traversal of the BST example two s

13、lides earlier It is sorted This is no coincidence As a general rule, if you output the keys (data) of the nodes of a BST using inorder traversal, the data comes out sorted in increasing order,CS 103,12,Other Kinds of Binary Trees (Full Binary Trees),Full Binary Tree: A full binary tree is a binary t

14、ree where all the leaves are on the same level and every non-leaf has two children The first four full binary trees are:,CS 103,13,Examples of Non-Full Binary Trees,These trees are NOT full binary trees: (do you know why?),CS 103,14,Canonical Labeling of Full Binary Trees,Label the nodes from 1 to n

15、 from the top to the bottom, left to right,Relationships between labels of children and parent:,2i,2i+1,i,CS 103,15,Other Kinds of Binary Trees (Almost Complete Binary trees),Almost Complete Binary Tree: An almost complete binary tree of n nodes, for any arbitrary nonnegative integer n, is the binar

16、y tree made up of the first n nodes of a canonically labeled full binary,1,1,2,1,2,3,4,5,6,7,1,2,1,2,3,4,5,6,1,2,3,4,1,2,3,4,5,CS 103,16,Depth/Height of Full Trees and Almost Complete Trees,The height (or depth ) h of such trees is O(log n) Proof: In the case of full trees, The number of nodes n is:

17、 n=1+2+22+23+2h=2h+1-1 Therefore, 2h+1 = n+1, and thus, h=log(n+1)-1 Hence, h=O(log n) For almost complete trees, the proof is left as an exercise.,CS 103,17,Canonical Labeling of Almost Complete Binary Trees,Same labeling inherited from full binary trees Same relationship holding between the labels

18、 of children and parents:,Relationships between labels of children and parent:,2i,2i+1,i,CS 103,18,Array Representation of Full Trees and Almost Complete Trees,A canonically label-able tree, like full binary trees and almost complete binary trees, can be represented by an array A of the same length

19、as the number of nodes Ak is identified with node of label k That is, Ak holds the data of node k Advantage: no need to store left and right pointers in the nodes save memory Direct access to nodes: to get to node k, access Ak,CS 103,19,Illustration of Array Representation,Notice: Left child of A5 (

20、of data 11) is A2*5=A10 (of data 18), and its right child is A2*5+1=A11 (of data 12). Parent of A4 is A4/2=A2, and parent of A5=A5/2=A2,6,15,8,2,11,18,12,20,27,13,30,15,8,20,2,11,30,27,13,6,10,12,1,2,3,4,5,6,7,8,9,10,11,CS 103,20,Adjustment of Indexes,Notice that in the previous slides, the node lab

21、els start from 1, and so would the corresponding arrays But in C/C+, array indices start from 0 The best way to handle the mismatch is to adjust the canonical labeling of full and almost complete trees. Start the node labeling from 0 (rather than 1). The children of node k are now nodes (2k+1) and (

22、2k+2), and the parent of node k is (k-1)/2, integer division.,CS 103,21,Application of Almost Complete Binary Trees: Heaps,A heap (or min-heap to be precise) is an almost complete binary tree where Every node holds a data value (or key) The key of every node is the keys of the children,Note: A max-h

23、eap has the same definition except that the Key of every node is = the keys of the children,CS 103,22,Example of a Min-heap,16,5,8,15,11,18,12,20,27,33,30,CS 103,23,Operations on Heaps,Delete the minimum value and return it. This operation is called deleteMin. Insert a new data value,Applications of

24、 Heaps:A heap implements a priority queue, which is a queue that orders entities not a on first-come first-serve basis, but on a priority basis: the item of highest priority is atthe head, and the item of the lowest priority is at the tailAnother application: sorting, which will be seen later,CS 103

25、,24,DeleteMin in Min-heaps,The minimum value in a min-heap is at the root! To delete the min, you cant just remove the data value of the root, because every node must hold a key Instead, take the last node from the heap, move its key to the root, and delete that last node But now, the tree is no lon

26、ger a heap (still almost complete, but the root key value may no longer be the keys of its children,CS 103,25,Illustration of First Stage of deletemin,16,8,15,11,18,12,20,27,33,30,16,8,15,11,18,12,20,27,33,30,16,8,15,11,18,12,20,27,33,30,CS 103,26,Restore Heap,To bring the structure back to its “hea

27、pness”, we restore the heap Swap the new root key with the smaller child. Now the potential bug is at the one level down. If it is not already the keys of its children, swap it with its smaller child Keep repeating the last step until the “bug” key becomes its children, or the it becomes a leaf,CS 1

28、03,27,Illustration of Restore-Heap,16,8,15,11,18,12,20,27,33,30,16,12,15,11,18,8,20,27,33,30,16,11,15,12,18,8,20,27,33,30,Now it is a correct heap,CS 103,28,Time complexity of insert and deletmin,Both operations takes time proportional to the height of the tree When restoring the heap, the bug moves

29、 from level to level until, in the worst case, it becomes a leaf (in deletemin) or the root (in insert) Each move to a new level takes constant time Therefore, the time is proportional to the number of levels, which is the height of the tree. But the height is O(log n) Therefore, both insert and del

30、etemin take O(log n) time, which is very fast.,CS 103,29,Inserting into a minheap,Suppose you want to insert a new value x into the heap Create a new node at the “end” of the heap (or put x at the end of the array) If x is = its parent, done Otherwise, we have to restore the heap: Repeatedly swap x

31、with its parent until either x reaches the root of x becomes = its parent,CS 103,30,Illustration of Insertion Into the Heap,In class,CS 103,31,The Min-heap Class in C+,class Minheap /the heap is implemented with a dynamic arraypublic:typedef int datatype;Minheap(int cap = 10)capacity=cap; length=0;p

32、tr = new datatypecap;datatype deleteMin( );void insert(datatype x);bool isEmpty( ) return length=0;int size( ) return length;private:datatype *ptr; / points to the arrayint capacity;int length;void doubleCapacity(); /doubles the capacity when needed ;,CS 103,32,Code for deletemin,Minheap:datatype Mi

33、nheap:deleteMin( )assert(length0);datatype returnValue = ptr0; length-; ptr0=ptrlength; / move last value to root elementint i=0;while (2*i+1ptr2*i+1) |(2*i+2ptr2*i+1 | ptriptr2*i+2) / “bug” still at least one childif (ptr2*i+1 = ptr2*i+2) / left child is the smaller childdatatype tmp= ptri; ptri=pt

34、r2*i+1; ptr2*i+1=tmp; /swapi=2*i+1; else / right child if the smaller child. Swap bug with right child.datatype tmp= ptri; ptri=ptr2*i+2; ptr2*i+2=tmp; / swapi=2*i+2; return returnValue; ;,CS 103,33,Code for Insert,void Minheap:insert(datatype x)if (length=capacity)doubleCapacity();ptrlength=x;int i=length;length+;while (i0 ,CS 103,34,Code for doubleCapacity,void Minheap:doubleCapacity()capacity = 2*capacity;datatype *newptr = new datatypecapacity;for (int i=0;ilength;i+)newptri=ptri;delete ptr;ptr = newptr; ;,