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Program of Binary tree:
Btree::Btree(int O) : itemsInContainer(0)
{
finishInit(O);
}
Btree::~Btree(void)
if( root != 0 )
delete root;
void Btree::flush( DeleteType dt )
int oldValue = ownsElements();
ownsElements( delObj(dt) );
itemsInContainer = 0;
root = 0;
ownsElements( oldValue );
int Btree::hasMember( Object& o ) const
if( !o.isSortable() )
ClassLib_error( __ENOTSORT );
if( root == 0 )
return 0;
else
Node* loc;
int idx;
return root->found(&(Sortable&)o, &loc, &idx) != NOOBJECT;
how to do 2, 4, 8 associativity
No. it's not required The C++ language guarantees that delete p will do nothing if p is equivalent to NULL. As you might get the test backwards, and as most testing methodologie
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Given a string, print all possible palindromic partitions using recursion
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