Deletion by merging in binary search tree

Hi there,

I'm reading about a technique known as deletion by merging in context of binary search trees. Here's the PDF:

http://www.cs.mun.ca/av/old/teaching...rees2_quad.pdf

and the code for 2 related functions: deleteByMerging and findAndDeleteByMerging, is on page 4-5. The book, from which the PDF comes from, says that:

Quote:

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Using search() in function findAndDeleteByMerging() is a treacherous simplification. search() returns a pointer to the node containing el. In findAndDeleteByMerging(), it is important to have this pointer stored specifically in one of the pointers of the node's parent. In other words, a caller to search() is satisfied if it can access the node from any direction, whereas findAndDeleteByMerging() wants to access it from either its parent's left or right pointer data member. Otherwise, access to the entire subtree having this node as its root would be lost.

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This quote pertains probably to this snippet on page 5:

Code:

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if(node != 0 && node->el == el)
    if(node == root)
        deleteByMerging(root);
    else if(prev->left == node)
        deleteByMerging(prev->left);
    else
        deleteByMerging(prev->right);

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I don't get it, why is it necessary to have 4 if-clauses? Wouldn't one, the outer, be sufficient? I mean, if prev->left == node or if prev->right == node then one way or another, we pass node in the call to deleteByMerging.

Wouldn't it be much simpler to just write:

Code:

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if(node != 0 && node->el == el)
    deleteByMerging(node);

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?

Quote:

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Originally Posted by Quentin026

I don't get it, why is it necessary to have 4 if-clauses?

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Well, it certainly looks suspicious. Still what you suggest is not equivalent to the original code.

In the original code if node is neither root nor prev->left then deleteByMerging(prev->right) will be called, whereas in your suggestion deleteByMerging(node) would be called.

It may be significant, I cannot say.

I didn't look at the original code, but I guess the parameter passed to deleteByMerging() is a non-const reference. In this case, even if prev->left == node, it does matter which of the two is passed: It determines which object gets modified by the function call.