Because t is passed by reference,the subtree is a,tomatically connected in Java

Encoding Code 3 of 9 in Java Because t is passed by reference,the subtree is a,tomatically connected
Because t is passed by reference,the subtree is a,tomatically connected
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Internal method to insert into a subtree x is the item to insert t is the node that roots the tree Set the new root Throw DuplicateItemException if x is already in t template <class Comparable> void BinarySearchTree<Cornparable>:: insert( const Comparable & x, Node * & t ) const i if ( t == NULL ) t = new Node ( x , NULL, NULL ) ; else if( x < t->element ) insert ( x, t->left ) ; else if( t->element < x ) insert( x, t->right ) ; else throw DuplicateItemException( ) ; 18 1
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Figure 1910 The recursive
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insert for the BinarySearchTree class
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Passing t by reference in the remove routines In effect we maintain the parent in the recursion stack
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to require us to keep track of the parent of the current node as we descend the tree But again, we can avoid the explicit use of a parent pointer by using recursion The code is shown in Figure 1 91 I If the tree t is empty, removeMin fails Otherwise if t has a left child, we recursively remove the minimum item in the left subtree via the recursive call at line 12 If we reach line 15, we know that we are currently at the minimum node, and thus t is the root of a subtree that has no left child If we set t to t- >right and then delete the node that t used to point at, t is now the root of a subtree that is missing its former minimum element That is what we do at lines 15-17 But doesn't that disconnect the tree The answer again is no If t was root, then, as t is passed by reference, root is changed to point at the new tree If t was not root,it is p->left , where p is t ' s parent at the time of the recursive call The change to t, being by reference, also changes p->left Thus the parent's left pointer points at t, and the tree is connected All in all it is a nifty maneuver-we have maintained the parent in the recursion stack rather than explicitly kept track of it in an iterative loop Having used this trick for the simple case, we can then adapt it for the general remove routine shown in Figure 1912 If the tree is empty, the remove is unsuccessful and we can throw an exception at line 10 If we do not have a match, we can recursively call remove for either the left or right subtree, as appropriate Otherwise, we reach line 15, indicating that we have found the node that needs to be removed
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Internal method to remove minimum item from a subtree
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/ / t is the node that roots the tree / / Set the new root / / Throws UnderflowException if L is empLy
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template <class Comparable> void BinarySearchTree<Comparable>:: removeMin( Node * & t ) const i if( r == NULL ) throw UnderflowExceptioni ) ; else if( t->lefr ! = NULL ) removeMin( t->left ) ; else
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Node *tmp = t; L = t->right; delete tmp; 1
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Figure 1911 The removeMin method for the BinarySearchTree class
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Internal method to remove from a subtree x is the item to remove, t is the node that roots the tree Set the new root Throw ItemNotFoundException is x is not in t templaLe <class Comparable> void BinarySearchTree<Comparable>:: remove( const Comparable & x , Node * & t ) consr I if( t == NULL ) throw ItemNotFol~ndException ) ; ( if ( x < L->element ) remove( x , t->left i ; else if( t->element < x ) remove ( x , t->right ) ; else if( t->left ! = NULL && t->right ! = NULL ) / / 2 children
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t->element = findMin( t->right )->element; removeMin( t->right ) ; / / Remove minimum 1 else / / One or zero children
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BinaryNode<Comparable> *oldNode = L ; t = ( t->left ! = NULL ) t->left : t->right; / / Reroot delete oldNode; / / delete old root 1
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