Tree Traversal: lterator Classes in Java

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Tree Traversal: lterator Classes
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/ / TreeIterator class interface; maintains "current position" //
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/ / CONSTRUCTION: with a tree to which the iterator is bound // / / ******************puBLIC OPERATIONS**********************
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First two are not virtual, last two are pure virtual boo1 isValid( ) - - > True if at valid position in tree Object retrieve( ) - - > Return item in current position void firsti ) - - > Set current position to first void advance ( ) - - > Advance **XX**************ERRORS****fi************************** 12 / / BadIterator is thrown for illegal access or advance 13 14 template <class Object> 15 class TreeIterator 16 i 17 public: 18 TreeIterator( const BinaryTreeiObject> & theTree ) 19 : root( theTreeroot ) , current( NULL ) { 1 20 virtual -TreeIterator( ) { 1 21 22 virtual void first ( ) = 0; 23 boo1 isValid( ) const { return current ! = NULL; } 24 const Object & retrieve( ) const; 25 virtual void advance( ) = 0; 26 protected: 27 28 const BinaryNode<Object> *root; 29 const BinaryNode<Object> *current; 30 1 ; 31 32 / / Return the item stored in the current position 33 / / Throw BadIterator exception if current position is invalid 34 template <class Object> 35 const Object & TreeIterator<Object>: :retrieve( ) const 36 ( 37 if( !isValid( ) ) 38 throw BadIterator( "Illegal retrieve" ) ; 39 return current->element; 40 }
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Figure 1825 The tree iterator abstract base class
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Each node is placed on the stack three times The third time off, the node is declared visited The other times, we simulate a recursive call
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When the stack is empty, every node has been visited
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Consequently, each node is placed on the stack three times during the course of the traversal If a node is popped from the stack a third time, we can mark it as the current node to be visited Otherwise, the node is being popped for either the first time or the second time In this case, it is not yet ready to be visited, so we push it back onto the stack and simulate a recursive call If the node was popped for a first time, we need to push the left child (if it exists) onto the stack Otherwise, the node was popped for a second time, and we push the right child (if it exists) onto the stack In any event, we then pop the stack, applying the same test Note that, when we pop the stack, we are simulating the recursive call to the appropriate child If the child does not exist and thus was never pushed onto the stack, when we pop the stack we pop the original node again Eventually, either the process pops a node for the third time or the stack empties In the latter case, we have iterated over the entire tree We initialize the algorithm by pushing a pointer to the root onto the stack An example of how the stack is manipulated is shown in Figure 1826 A quick summary: The stack contains nodes that we have traversed but not yet completed When a node is pushed onto the stack, the counter is 1 : 2, or 3 as follows: 1: If we are about to process the node's left subtree, 2: if we are about to process the node's right subtree, or 3: if we are about to process the node itself
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Figure 1826 Stack states during postorder traversal
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Tree Traversal: Iterator Classes
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Let us trace through the postorder traversal We initialize the traversal by ' pushing root a onto the stack The first pop visits a This is a s first pop, so it is placed back on the stack, and we push its left child b, onto the stack Next b is popped It is b's first pop, so it is placed back on the stack Normally, b's left child would then be pushed, but b has no left child, so nothing is pushed Thus the next pop reveals b for the second time, b is placed back on the stack, and its right child, d,is pushed onto the stack The next pop produces d for the first time, and d is pushed back onto the stack No other push is performed because d has no left child Thus d is popped for the second time and is pushed back, but as it has no right child, nothing else is pushed Therefore the next pop yields d for the third time, and d is marked as a visited node The next node popped is b, and as this pop is b's third, it is marked visited Then a is popped for the second time, and it is pushed back onto the stack along with its right child, cNext, c is popped for the first time, so it is pushed back, along with its left child, eNow e is popped, pushed, popped, pushed, and finally popped for the third time (typical for leaf nodes) Thus e is marked as a visited node Next, c is popped for the second time and is pushed back onto the stack However, it has no right child, so it is immediately popped for the third time and marked as visited Finally, a is popped for the third time and marked as visited At this point, the stack is empty and the postorder traversal terminates The POStorder class is implemented directly from the algorithm described previously and is shown in Figure 1827 The StNode class represents the objects placed on the stack It contains a pointer to a node and an integer that stores the number of times the item has been popped from the stack An StNode object is always initialized to reflect the fact that it has not yet been popped from the stack (We use a Stack class from 16) The Postorder class is derived from TreeIterator and adds an internal stack to the inherited data members The Postorder class is initialized by initializing the TreeIterator data members and then pushing the root onto the stack This process is illustrated in the constructor at lines 34 to 39 Then first is implemented by clearing the stack, pushing the root, and calling advance Figure 1828 implements advanceIt follows the outline almost verbatim Line 7 tests for an empty stack If the stack is empty, we have completed the iteration and can set current to NULL and return (If current is already NULL,we have advanced past the end, and an exception is thrown) Otherwise, we repeatedly perform stack pushes and pops until an item emerges from the stack for a third time When this happens, the test at line 19 is successful and we can return Otherwise, at line 25 we push the node back onto the stack (note that the timesPopped component has already been incremented at line 19) We then implement the recursive call If the
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An StNode stores a pointer to a node and a count that tells how many times it has already been popped
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