我在AvlTree类中实现了一个类迭代器。我的AvlTree节点如下:
struct AvlNode
{
Comparable element;
list<int> lines; //line occurrences
bool flag; //checks validity
AvlNode *left;
AvlNode *right;
AvlNode *parent; //parent pointer
int height;
AvlNode( const Comparable & theElement, AvlNode *lt, AvlNode *rt, AvlNode *pt,
int h = 0, bool b = true )
: element( theElement ), left( lt ), right( rt ), parent( pt ), height( h ), flag( b ) { }
};
我的迭代器如下:
class iterator
{
protected:
friend class AvlTree<Comparable>;
AvlNode * node;
AvlNode * findInOrderSuccessor(AvlNode * & t)
{
AvlNode * temp;
//node has a right child
// so successor is leftmost node of right subtree
if(t->right != NULL)
{
temp = t->right; //go right
//go all the way left
while(temp->left != NULL)
{
temp = temp->left;
}
return temp;
}
//node has no right child
//if we are someone's left child, go up one
if(t->parent->left == t)
{
//return your parent
temp = t->parent;
return temp;
}
//if we are someone's right child, go up until the current node
//is someone's left child, then go up one more
temp = t->parent;
while(temp->parent->left != temp)
{
temp = temp->parent; //go up
}
//return your parent
temp = t->parent;
return temp;
}
public:
iterator(AvlNode * p) : node(p)
{ }
//overload * to make *iterator return the element of its node
Comparable & operator*()
{ return node->element; }
iterator operator++ (int) //postfix operator
{
node = findInOrderSuccessor(node);
return iterator(node);
}
// == comparison overload
bool operator==(iterator rhs)
{ return node == rhs.node; }
// != comparison overload
bool operator!=(iterator rhs)
{ return !(*this == rhs); }
};
我的AvlTree还有一个开始和结束迭代器作为公共成员:
//begin iterator points to leftmost node
iterator begin()
{ //return pointer to leftmost node
AvlNode *temp = root;
while(temp->left != NULL)
temp = temp->left;
return iterator(temp);
}
//end iterator points to one after rightmost node
iterator end()
{ //return NULL right pointer of rightmost node
AvlNode * temp = root;
while(temp->right != NULL)
temp = temp->right;
return iterator( temp->right );
}
我的问题是当我尝试在main中运行以下内容时:
for(AvlTree<string>::iterator itr = tree.begin(); itr != (tree.end()); itr++)
cout << *itr << endl;
不是按顺序输出字符串树中的所有单词,而是获得树中第一个顺序项的无限循环。我似乎无法弄清楚为什么它不会超过第一个项目。
答案 0 :(得分:1)
以下迭代代码有效(来自my AVL tree;将left
和right
替换为link[0]
和link[1]
):
BAVLNode * BAVL_GetFirst (const BAVL *o)
{
if (!o->root) {
return NULL;
}
BAVLNode *n = o->root;
while (n->link[0]) {
n = n->link[0];
}
return n;
}
BAVLNode * BAVL_GetNext (const BAVL *o, BAVLNode *n)
{
if (n->link[1]) {
n = n->link[1];
while (n->link[0]) {
n = n->link[0];
}
} else {
while (n->parent && n == n->parent->link[1]) {
n = n->parent;
}
n = n->parent;
}
return n;
}
就您的代码而言,首先,end()
不需要找到最右边的节点才能返回iterator(NULL)
;它可以在不看树的情况下返回。
然而,算法中的实际错误似乎在这里:
temp = t->parent;
WRONG: while(temp->parent->left != temp)
{
temp = temp->parent; //go up
}
//return your parent
WRONG: temp = t->parent;
return temp;
}
我标记的第一行可能会尝试NULL指针取消引用,应该更改为:
while(temp->parent && temp->parent->left != temp)
第二个:
temp = temp->parent;
另外,您可能已经从我的代码中注意到以下内容现在非常丰富;它可以被移除,并且(固定的)剩余代码将以完全相同的方式处理它。它也遭受了我在上面指出的相同的NULL指针解除引用。
//if we are someone's left child, go up one
if(t->parent->left == t)
{
//return your parent
temp = t->parent;
return temp;
}