我遇到的问题是我的树只能成功删除一个节点一次。当我尝试删除更多时，它会因分段错误（核心转储）错误而崩溃。我似乎无法确定故障的确切位置，但它应该是我在内存分配和释放方面做错了，我只是无法看到它。
这是我的插入，删除，删除帮助器和叶子（因为它被称为内部删除）：
插入：
/*********************************************************
Function: insert
Arguments: const T &x (a T variable)
Returns: void
Notes: If the tree is empty, the function will set the node
as the root. Otherwise, it will look for the appropiate
place (in terms of a BST) to place it as long as its data
is not already existent inside the tree.
*********************************************************/
template<class T> void binSTree<T>::insert(const T &x)
{
BinTreeNode *newnode, *current, *parentOfCurrent; /* newnode will be the inserted node, current and parentOfCurrent will be traversing the while loop to find an appropiate space to insert */
newnode = new BinTreeNode; /* allocates storage space */
newnode->info = x; /* sets newnode's data equal to x */
newnode->left = newnode->right = NULL; /* sets newnode's left and right children equal to NULL */
if (root == NULL) { /* if the tree is empty, it will place newnode on the root */
root = newnode;
cout << "added to root\n";
return; }
current = root; /* sets current at the root */
while(current != NULL) { /* while current is not NULL, it will search for a place to insert the new node */
parentOfCurrent = current; /* sets the parent equal to current */
if(current->info == x) /* if the variable is found inside the tree, then it will error and exit */
{
cout << x << " already exists in tree. Duplicates not allowed!\n";
return;
}
else if(current->info > x) /* otherwise, if the variable is less than the data on the tree currently, it will move left */
current = current->left;
else /* otherwise, if the variable is greater than the data on the tree currently, it will move right */ current = current->right;
}
/* the new node is placed where current would be in */
if (parentOfCurrent->info > x)
parentOfCurrent->left = newnode;
else
parentOfCurrent->right = newnode;
}
删除和privRemove：
template <class T> bool binSTree<T>::remove(const T &x)
{
BinTreeNode *current, *parentOfCurrent; /* current for tranversing the tree in the while loop and parentofCurrent to help */
bool found = false; /* the booleans that is to be returned */
if (root == NULL) { /* Checks if the tree is empty, if it is, it returns found (false) and exits */
return found;
}
current = root; /* sets current equal to root */
while(current != NULL) /* loops repeats until current is equal to NULL */
{
if(current->info == x) { /* if the current data from the tree is equal tox, then it breaks the loop and sets found equal to true */
found = true;
break;
}
else { /* otherwise, it will search for the next place to go in the tree */
parentOfCurrent = current; /* parentOfCurrent is set to current before current is set to one of its children */
if(current->info > x) /* if x is less than the data on current, then it will move left */
current = current->left;
else /* if x is greater than the data on current, then it will move right */
current = current->right;
}
}
if(found == true) { /* if it was found, it will pass current via the parent to the private remove function */
if (current == root)
privRemove(root);
else if (parentOfCurrent->info > x) { /* if current is to the left of the parent */
found = leaf( parentOfCurrent->left );
if(found == true)
privRemove(parentOfCurrent->left);
}
else { /* if current is to the right of the parent */
found = leaf( parentOfCurrent->left );
if(found == true)
privRemove(parentOfCurrent->right);
}
}
return found;
}
/*********************************************************
Function: privRemove
Arguments: BinTreeNode* &node (the node that is to be deleted)
Returns: void
Notes: This private remove function will take in the node that
is provided by the public remove function, and, after ensuring
that the node passed is not NULL, then it will delete it.
*********************************************************/
template <class T> void binSTree<T>::privRemove(BinTreeNode* &node)
{
BinTreeNode *temp; /* initializes temp in order to hold the information of node */
if(root != NULL ) /* if the node is not NULL */
{
temp = node;
delete node;
node = temp;
}
}
叶：
/*********************************************************
Function: leaf
Arguments: BinTreeNode* node
Returns: boolean
Notes: This function will check if the node in the argument
is a leaf by checking if both of its children are NULL. If
they are, it returns true. Otherwise, it returns false.
*********************************************************/
template <class T> bool binSTree<T>::leaf(BinTreeNode* node) const
{
bool isLEaf = false; /* the boolean variable it is to return */
if(node->left == NULL && node->right == NULL) /* checks if the node has no children */
isLEaf = true; /* if it does, it sets isLEaf equal to true */
return isLEaf; /* after going through the if statement, it returns the variable isLeaf */
}
应该注意，我的节点是在类头中声明的结构，并且变量root被声明为BinTreeNode * root形式的受保护变量;它在binSTree构造函数中初始化为NULL。此外，我只允许删除叶子。
答案 0 :(得分：1)
在我看来你做错了是在privRemove方法中...... 想象一下，你试图删除一个节点，并在该节点上调用privRemove。
使用：
delete node;
释放内存节点指向。
然后你将节点重新分配给之前的值：
node=temp;
因此它现在指向与之前相同的位置（因为您通过引用传递了它而在函数的内部和外部）。
现在在下一次删除时，你最终会遇到同一个节点并测试它是否为NULL，但它不是因为你给它一个值所以它指向它之前的内存（node = temp），所以节点指向内存中某个非NULL的位置，所以你会认为它是一个合法的节点但它不是，它指向已经释放的内存。当你在上面调用一个方法时... bhumm！