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Why wouldnt we just free the root. Doesnt it free all the memories that are bound to it ? I may be wrong. I even know I am wrong but , so even we free the root , other memorial places will still be filled?Code:void destroy_tree(struct node *leaf)
{
if( leaf != 0 )
{
destroy_tree(leaf->left);
destroy_tree(leaf->right);
free( leaf );
}
}
No AFAIK it won't. It will only free the root node.
No, it doesn't work like that. C has no concept of a "tree". You asked to allocate some memory, C doesn't know/care how you're going to use that memory. If you were to free the root, you'd simply free that one element, leaving all the other elements still allocated and no way to get to them. That's called a memory leak ;)
QuantumPete
Yes. If you do not explicitly free the nodes, they will continue to fill memory. For newer languages, like Java, this is generally not the case.
It looks good to me.
I think the OP was asking *why* you need to go through each node and free it, I'm sure the code is out of a book somewhere.Quote:
Originally Posted by Salem
QuantumPete
Yes that is true . Besides I dont have any problem with the code. I have seen it somewhere. You have already answered my question.Thanks to all.
memory 4 the each node will be allocated separately and if u are freeing only root node,then only node is getting freed and memory 4 the others node is still allocatedQuote:
Originally Posted by ozumsafa