因此,我正在研究BST,构建删除工具。修改递归中的调用指针
我的代码序列似乎工作正确 - 保存没有更新父或根,并设置将它发送到删除的节点的地址为NULL的指针。
我传递了一个指针指向我的Erase和RemoveNode函数中的一个指针,以便直接影响实际导致递归调用的左,右或根数据成员。在遍历代码时,它将removeN函数中的* N设置为NULL,但这不会在调用对象的数据中反映出来。我在使用指针指针方法时不正确吗?如果是这样,是否有一种方法可以递归地删除并能够修改先前的节点,如果链接被销毁?
节点结构:
struct tNode
{
tNode(int n)
{
data = n;
left = NULL;
right = NULL;
}
//Works, cleans all linked objects.
//Must remember to null links when removing wanted nodes
~tNode(void)
{
//cout << "Deleting " << data << endl;
delete left;
delete right;
}
// Data members
int data;
tNode* left;
tNode* right;
};
橡皮擦功能递归在树:
void BinSearchTree::Erase(int n, tNode** N)
{
tNode* node = *N;
if (root)
{
if (node->data > n) // post order, to avoid moving many times over
{
if (node->left)
{
Erase(n, &node->left);
}
}
else
{
if (node->right)
{
Erase(n, &node->right);
}
}
if (node->data == n)
{
RemoveNode(&node);
}
}
}
而且的removeNode函数来处理实际删除:
void BinSearchTree::RemoveNode(tNode** N)
{
tNode* node = *N;
if (!node->left && !node->right) // is leaf
{
delete node; // remove node
size--;
*N = NULL; // null pointer for above node/structure
}
else if (!node->left) // right child
{
tNode* temp = node->right; // to strip out copied node when finished
node->data = node->right->data; // copy right node into current node
node->right = node->right->right;
node->left = node->right->left;
temp->right = NULL; // NULL because node destructor is recursive
temp->left = NULL; // ^^
delete temp;
size--;
}
else if (!node->right) // left child
{
tNode* temp = node->left; // to strip out copied node when finished
node->data = node->left->data; // copy left node into current node
node->right = node->left->right;
node->left = node->left->left;
temp->right = NULL; // NULL because node destructor is recursive
temp->left = NULL; // ^^
delete temp;
size--;
}
else // 2 children
{
tNode* temp = node->right; // find ideal child -> left-most right child
tNode* parent = NULL; // keep track of owner of ideal child
while (temp->left)
{
parent = temp;
temp = temp->left;
}
node->data = temp->data; // copy ideal child to root
if (parent)
{
parent->left = temp->right; // case that left-most child has right child of it's own
}
RemoveNode(&temp);
size--;
}
}
您是否有真正的问题? –
@KerrekSB对不起,我一定领先于自己。感谢您指出。 – DivinusVox