Question

Can someone tell me what the following lines do

else if ( obj < retrieve() ) 
{
    return ( left() == 0 ) ? 0 : left()->erase( obj, left_tree );
} 
else 
{
    return ( right() == 0 ) ? 0 : right()->erase( obj, right_tree );
}

in the code block below:

    template <typename Comp>
    int Binary_search_node<Comp>::erase( Comp const &obj, Binary_search_node<Comp> *&ptr_to_this) 
{
    if ( obj == retrieve() ) {
        if ( leaf() ) { // leaf node
            ptr_to_this= 0;
            delete this;
        } 
        else if ( left() != 0 && right() != 0 ) { // full node
            element= right()->front();
            right()->erase( retrieve(), right_tree );
        } 
        else { // only one child
            ptr_to_this= ( left() != 0 ) ? left() : right();
            delete this;
        }
        return 1;
    } 
    else if ( obj < retrieve() ) {
        return ( left() == 0 ) ? 0 : left()->erase( obj, left_tree );} 
    else {
        return ( right() == 0 ) ? 0 : right()->erase( obj, right_tree );}
}

Extra Information:

1)

front() -- finds the minimum objects

Implementation:

template <typename Comp>
Comp Binary_search_node<Comp>::front() const 
{
    return( left() == 0 ) ?retrieve() :left()->front();
}

2)

left()  -- returns pointer to left subtree

3)

right() -- returns pointer to right subtree

4)

*ptr_to_this points to current object (same location as what *this points to)

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I have an idea of what the lines do, but I am not 100% sure thus I wanted to confirm. Please note that this erase() function is for a binary search tree. Thanks!

Answer 1

erase appears to be implemented recursively. At each stage, we test whether the object to erase is equal to the current object, or whether we need to go down into the left or right child.

If the child that we want to go into does not exist (left() == 0 or right() == 0), then we cannot erase the object (because it’s not in the tree), so we return 0. Otherwise, we recurse into the child function, and return whatever it returns.