在C ++中的删除操作中出现段错误

Question

i am writing a code for a c++ assignment, it is a dictionary implementation using a binary search tree. My code compiles but when i try to "remove" i get a seg Fault. any ideas why did might be happening. Thanks

Here is my code

// this function calls the deleteNode function where the deletion is done
void BST::deleteContent(string *word)
{
    deleteNode(word, root);
}
// a helper fuuntion for the deletecontent function
//uses recursion to find the node to be deleted
void BST::deleteNode(const string *word, Node *&nodePtr)
{
    if(word < nodePtr->word)
        deleteNode(word, nodePtr->left);
    else if(word > nodePtr->word)
        deleteNode(word, nodePtr->right);
    else
        makeDeletion(nodePtr);
}
// a helper function for the deleteNode function
void BST::makeDeletion(Node *&nodePtr)
{
    Node *tempNodePtr;

    if(nodePtr == NULL)
        cout<< "cannot delete empty node. \n";
    // if node has no right child
    else if (nodePtr->right == NULL)
        {
            tempNodePtr = nodePtr;
            nodePtr = nodePtr->left; // reattach child
            delete tempNodePtr;
        }
    else if(nodePtr-> left == NULL)
        {
            tempNodePtr = nodePtr;
            nodePtr = nodePtr->right; // reattach child
            delete tempNodePtr;
        }
    // if node has 2 children
    else
        {
            tempNodePtr = nodePtr->right;
            while (tempNodePtr->left)
                tempNodePtr = tempNodePtr->left;
            tempNodePtr->left = nodePtr->left;
            tempNodePtr = nodePtr;
            nodePtr = nodePtr->right;
            delete tempNodePtr;
        }
}

EDIT :

Thank you all!! From your post i realised it was a good idea to check if the node was the last and had no children. I added this check in deleteNode

    if((nodePtr->left) && word < nodePtr->word)
    {
        do something
    }

i did the same for the right it worked and did not throw any errors or seg faults. Thanks so much!!!!

Answer 1

Case 1: Empty tree:

Suppose your your tree is empty: root will be nullptr . So deleteContent() will call deleteNode() with argument nullptr for nodePtr .

The first thing you do there is compare word with nodePtr->word without first checking that nodePtr is not nullptr . THere you have a first case of segmenation fault !

Case 2: Delete a word which is not in the tree:

In this case, deleteNode() will be called recursively until reaching a leaf node with no descendent. As the searched word does not exist in the tree, it's either geater or lesser than nodePtr->word, but never equal. deleteNode() will then call itself, again passing argument nullptr for nodePtr , as in case 1. Again you'll have a segmentation fault !

Solution to case 1 and 2: Control nullptr in deleteNode():

void BST::deleteNode(const string *word, Node *&nodePtr)
{
    if (nodePtr==nullptr) 
        cout << word << " not found in the tree\n";
    else if (word < nodePtr->word)
        ...   // the rest as in your original function 
}

makeDeletion() should now be called by deleteNode() if and only if nodePtr is not null and word==nodePtr->word . Get rid of the first if() which should no longer be true in any case. May be replace it with an assert at to verify the invariant.

Case 3: Delete a word in the tree :

All the three cases seem to work (even leaf nodes with two null pointers), at least if I look at my drawing of your data structure.

However I'd suggest to verify Node::~Node() : in all cases, you reattach the children and then you delete the old node ( temNodePtr ) without having set its children to nullptr . So I wonder whether ~Node() just destroys the node without taking care of its children (then makeDeletion() should work) or if its a recursive destructor, deleteing the node and its children (then makeDeletion() would not work because, you would delete the nodes that you've just reattached, without noticing it, thus creating a seg.fault at the first occasion).

By the way, nullptr would perhap's be be more appropriate than NULL for pointers, even if NULL would work properly as well.