重新平衡时如何修复 AVL 删除操作中的分段错误?
How to fix segmentation fault in AVL deletion operation when rebalancing?
我正在实现一个 AVL 树,我的搜索和插入功能正常工作,但我的删除功能出现分段错误。我之前已经正确实现了 BST 树,所以我知道问题在于树的重新平衡,而不是节点的初始删除。
因为我的插入操作与重新平衡一起工作,所以我也知道问题不在于旋转函数本身。
我尝试了不同的策略,例如在每个节点上保持平衡因子,并尝试实施我在网上找到的其他源代码,但我总是以分段错误告终,而且真的找不到位置。如果有任何帮助,我将不胜感激。
class AVL
{
public:
AVL();
Node* insert(int num);
Node* search(int num);
Node* remove(int num);
void print();
void comparisonPrint();
private:
int comparisonCount;
Node* root;
int max(int a, int b);
int getHeight(Node* t);
int getBalance(Node* t);
Node* insert(Node* &t, int num);
Node* rotateWithLeftChild(Node* t);
Node* rotateWithRightChild(Node* t);
Node* doubleRotateWithLeftChild(Node* t);
Node* doubleRotateWithRightChild(Node* t);
Node* search(Node* t, int num);
Node* removeMin(Node* parent, Node* node);
Node* remove(Node* &t, int num);
void print(Node* t);
//print
};
int AVL::max(int a, int b)
{
return (a > b)? a : b;
}
int AVL::getHeight(Node* t)
{
return (t == NULL) ? 0 : t->height;
}
int AVL::getBalance(Node* t)
{
if(t == NULL)
return 0;
return getHeight(t->leftChild) - getHeight(t->rightChild);
}
//helper function for remove - finds min
Node* AVL::removeMin(Node* parent, Node* node) //removes node, but does not delete - returns ptr instead
{
if(node != NULL)
{
if(node->leftChild != NULL) //go to leftmost child in right subtree
return removeMin(node, node->leftChild);
else //min val
{
parent->leftChild = node->rightChild;
return node;
}
}
else //subtree empty - incorrect use of function
return NULL;
}
Node* AVL::remove(Node* &t, int num)
{
cout << num;
if(t != NULL)
{
if(num > t->key)
{
comparisonCount++;
remove(t->rightChild, num);
}
else if(num < t->key)
{
comparisonCount++;
remove(t->leftChild, num);
}
else if(t->leftChild != NULL && t->rightChild != NULL)
{
comparisonCount++;
//2 children
Node* minRightSubtree = removeMin(t, t->rightChild);
t->key = minRightSubtree->key;
delete minRightSubtree;
}
else
{
comparisonCount++;
//0 or 1 child
Node* temp = t;
if(t->leftChild != NULL)
t = t->leftChild;
else if(t->rightChild != NULL)
t = t->rightChild;
delete temp;
}
//update height
t->height = max(getHeight(t->leftChild), getHeight(t->rightChild)) + 1;
int balance = getBalance(t);
if(balance > 1 && getBalance(t->leftChild) >= 0)
return rotateWithRightChild(t);
if(balance > 1 && getBalance(t->leftChild) < 0)
{
t->leftChild = rotateWithLeftChild(t->leftChild);
return rotateWithRightChild(t);
}
if(balance < -1 && getBalance(t->rightChild) <= 0)
return rotateWithLeftChild(t);
if(balance < -1 && getBalance(t->rightChild) > 0)
{
t->rightChild = rotateWithRightChild(t->rightChild);
return rotateWithLeftChild(t);
}
}
return t;
}
所以我需要删除函数来删除指定的节点并在必要时使用适当的旋转重新平衡树。但是,每当我尝试在我的 main 中调用该函数时,我都会遇到分段错误。谢谢
您的代码有两个问题。首先是removeMin
函数和remove
函数中的else if
部分,当要删除的节点有两个子节点时。
removeMin
函数的基本目标应该是找到要删除的节点的顺序后继,在您的情况下是 t
。考虑 t
有 2 个子节点(均为叶节点)的情况,然后您的 removeMin
函数会将 t->leftChild
设置为 t->rightChild->rightChild
,这是错误的 NULL
。此外,树的重组应该在 remove
内完成,因此 removeMin
变为:
Node* AVL::removeMin(Node* node) // returns inorder successor of 't'
{
if(node->left == NULL)
return node;
return removeMin(node->left);
}
来到remove
函数,我们将t->key
重置为minRightSubtree->key
,现在要删除的节点是minRightSubtree
。但是请注意,从节点 t
到节点 minRightSubtree
,链中键的顺序发生了变化。 t->key
小于minRightSubtree
之前节点的所有key。因此你不能只是 delete minRightSubtree
,你必须在节点 minRightSubtree
上调用 remove
函数,这将负责重组这条链。您还可以从递归堆栈中获得一些帮助,以在 deletion/rotation:
之后为当前节点 t
获取正确的子节点
Node* AVL::remove(Node* &t, int num)
{
if (t == NULL)
return NULL;
if (num > t->key)
t->rightChild = remove(t->rightChild, num);
else if (num < t->key)
t->leftChild = remove(t->leftChild, num);
else if (t->leftChild != NULL && t->rightChild != NULL)
{
//2 children
Node* minRightSubtree = removeMin(t->rightChild);
t->key = minRightSubtree->key;
t->rightChild = remove(t->rightChild, minRightSubtree->key);
}
else
{
//0 or 1 child
Node* temp = t;
if (t->leftChild != NULL)
t = t->leftChild;
else if (t->rightChild != NULL)
t = t->rightChild;
if(temp == t)
t = NULL;
delete temp;
}
if (t == NULL) // this case was added since there is a possibility of deleting 't'
return NULL;
//update height
t->height = max(getHeight(t->leftChild), getHeight(t->rightChild)) + 1;
int balance = getBalance(t);
if (balance > 1 && getBalance(t->leftChild) >= 0)
return rotateWithRightChild(t);
if (balance > 1 && getBalance(t->leftChild) < 0)
{
t->leftChild = rotateWithLeftChild(t->leftChild);
return rotateWithRightChild(t);
}
if (balance < -1 && getBalance(t->rightChild) <= 0)
return rotateWithLeftChild(t);
if (balance < -1 && getBalance(t->rightChild) > 0)
{
t->rightChild = rotateWithRightChild(t->rightChild);
return rotateWithLeftChild(t);
}
return t;
}
我假设你更新高度和平衡根子树的代码是正确的,因为我忘记了它的理论并且需要修改。
我正在实现一个 AVL 树,我的搜索和插入功能正常工作,但我的删除功能出现分段错误。我之前已经正确实现了 BST 树,所以我知道问题在于树的重新平衡,而不是节点的初始删除。
因为我的插入操作与重新平衡一起工作,所以我也知道问题不在于旋转函数本身。
我尝试了不同的策略,例如在每个节点上保持平衡因子,并尝试实施我在网上找到的其他源代码,但我总是以分段错误告终,而且真的找不到位置。如果有任何帮助,我将不胜感激。
class AVL
{
public:
AVL();
Node* insert(int num);
Node* search(int num);
Node* remove(int num);
void print();
void comparisonPrint();
private:
int comparisonCount;
Node* root;
int max(int a, int b);
int getHeight(Node* t);
int getBalance(Node* t);
Node* insert(Node* &t, int num);
Node* rotateWithLeftChild(Node* t);
Node* rotateWithRightChild(Node* t);
Node* doubleRotateWithLeftChild(Node* t);
Node* doubleRotateWithRightChild(Node* t);
Node* search(Node* t, int num);
Node* removeMin(Node* parent, Node* node);
Node* remove(Node* &t, int num);
void print(Node* t);
//print
};
int AVL::max(int a, int b)
{
return (a > b)? a : b;
}
int AVL::getHeight(Node* t)
{
return (t == NULL) ? 0 : t->height;
}
int AVL::getBalance(Node* t)
{
if(t == NULL)
return 0;
return getHeight(t->leftChild) - getHeight(t->rightChild);
}
//helper function for remove - finds min
Node* AVL::removeMin(Node* parent, Node* node) //removes node, but does not delete - returns ptr instead
{
if(node != NULL)
{
if(node->leftChild != NULL) //go to leftmost child in right subtree
return removeMin(node, node->leftChild);
else //min val
{
parent->leftChild = node->rightChild;
return node;
}
}
else //subtree empty - incorrect use of function
return NULL;
}
Node* AVL::remove(Node* &t, int num)
{
cout << num;
if(t != NULL)
{
if(num > t->key)
{
comparisonCount++;
remove(t->rightChild, num);
}
else if(num < t->key)
{
comparisonCount++;
remove(t->leftChild, num);
}
else if(t->leftChild != NULL && t->rightChild != NULL)
{
comparisonCount++;
//2 children
Node* minRightSubtree = removeMin(t, t->rightChild);
t->key = minRightSubtree->key;
delete minRightSubtree;
}
else
{
comparisonCount++;
//0 or 1 child
Node* temp = t;
if(t->leftChild != NULL)
t = t->leftChild;
else if(t->rightChild != NULL)
t = t->rightChild;
delete temp;
}
//update height
t->height = max(getHeight(t->leftChild), getHeight(t->rightChild)) + 1;
int balance = getBalance(t);
if(balance > 1 && getBalance(t->leftChild) >= 0)
return rotateWithRightChild(t);
if(balance > 1 && getBalance(t->leftChild) < 0)
{
t->leftChild = rotateWithLeftChild(t->leftChild);
return rotateWithRightChild(t);
}
if(balance < -1 && getBalance(t->rightChild) <= 0)
return rotateWithLeftChild(t);
if(balance < -1 && getBalance(t->rightChild) > 0)
{
t->rightChild = rotateWithRightChild(t->rightChild);
return rotateWithLeftChild(t);
}
}
return t;
}
所以我需要删除函数来删除指定的节点并在必要时使用适当的旋转重新平衡树。但是,每当我尝试在我的 main 中调用该函数时,我都会遇到分段错误。谢谢
您的代码有两个问题。首先是removeMin
函数和remove
函数中的else if
部分,当要删除的节点有两个子节点时。
removeMin
函数的基本目标应该是找到要删除的节点的顺序后继,在您的情况下是 t
。考虑 t
有 2 个子节点(均为叶节点)的情况,然后您的 removeMin
函数会将 t->leftChild
设置为 t->rightChild->rightChild
,这是错误的 NULL
。此外,树的重组应该在 remove
内完成,因此 removeMin
变为:
Node* AVL::removeMin(Node* node) // returns inorder successor of 't'
{
if(node->left == NULL)
return node;
return removeMin(node->left);
}
来到remove
函数,我们将t->key
重置为minRightSubtree->key
,现在要删除的节点是minRightSubtree
。但是请注意,从节点 t
到节点 minRightSubtree
,链中键的顺序发生了变化。 t->key
小于minRightSubtree
之前节点的所有key。因此你不能只是 delete minRightSubtree
,你必须在节点 minRightSubtree
上调用 remove
函数,这将负责重组这条链。您还可以从递归堆栈中获得一些帮助,以在 deletion/rotation:
t
获取正确的子节点
Node* AVL::remove(Node* &t, int num)
{
if (t == NULL)
return NULL;
if (num > t->key)
t->rightChild = remove(t->rightChild, num);
else if (num < t->key)
t->leftChild = remove(t->leftChild, num);
else if (t->leftChild != NULL && t->rightChild != NULL)
{
//2 children
Node* minRightSubtree = removeMin(t->rightChild);
t->key = minRightSubtree->key;
t->rightChild = remove(t->rightChild, minRightSubtree->key);
}
else
{
//0 or 1 child
Node* temp = t;
if (t->leftChild != NULL)
t = t->leftChild;
else if (t->rightChild != NULL)
t = t->rightChild;
if(temp == t)
t = NULL;
delete temp;
}
if (t == NULL) // this case was added since there is a possibility of deleting 't'
return NULL;
//update height
t->height = max(getHeight(t->leftChild), getHeight(t->rightChild)) + 1;
int balance = getBalance(t);
if (balance > 1 && getBalance(t->leftChild) >= 0)
return rotateWithRightChild(t);
if (balance > 1 && getBalance(t->leftChild) < 0)
{
t->leftChild = rotateWithLeftChild(t->leftChild);
return rotateWithRightChild(t);
}
if (balance < -1 && getBalance(t->rightChild) <= 0)
return rotateWithLeftChild(t);
if (balance < -1 && getBalance(t->rightChild) > 0)
{
t->rightChild = rotateWithRightChild(t->rightChild);
return rotateWithLeftChild(t);
}
return t;
}
我假设你更新高度和平衡根子树的代码是正确的,因为我忘记了它的理论并且需要修改。