好的，这里是 splay 算法，如果你想检查的话。
这是我的伸展功能：
template <class WA>
void SplayTree<WA>::Do_Splay(SplayNODE<WA> *temp) //temp is the node which is to be splayed
{
if (temp==root) //if temp is root then
{ return; } //Do Nothing
else if (root->left==temp || root->right==temp) //else if temp is a child of root
{
if (root->left==temp) //if temp is left child of root then
{
Zig(root); //perform ZIG
}
else if (root->right==temp) //else if temp is right child of root then
{
Zag(root); //perform ZAG
}
}
else //if temp is some node lower in tree then
{
SplayNODE<WA> *father = findfather(temp, root);
SplayNODE<WA> *grandfather = findfather(father, root);
//cout<<"\n\tf = "<<father->key<<"\tgf = "<<grandfather->key;
////--------------------------------------------------------------------//-------////
if ( grandfather->left == father) //if father itself is left child
{
if(father->left == temp) //if temp is left child of father then
{ //CASE = ZIG ZIG
cout<<"\tZig(father)---Zag(grandfather)";
Zig(grandfather);
Zig(father);
}
else if ( father->right == temp ) //if temp is right child of father then
{ //CASE = ZIG ZAG
cout<<"\tZig(father)---Zag(grandfather)";
Zig(father);
Zag(grandfather);
}
}
//--------------------------------------------------------------------//-------////
if (grandfather->right == father) //if father itself is right child
{
if(father->right == temp)
{ //CASE = ZAG ZAG
cout<<"\tZag(grandfather)---Zag(father)";
Zag(grandfather);
Zag(father);
}
else if (father->left == temp)
{ //CASE = ZAG ZIG
cout<<"\tZag(father)---Zig(grandfather)";
Zag(father);
Zig(grandfather);
}
}
////--------------------------------------------------------------------//-------////
}
}
以下是 Zig (Single Rotate Left) & Zag (Single Rotate Right) 的方法。
template <class WA>
void SplayTree<WA>::Zig(SplayNODE<WA> * & k2) //k2 is temp node where imbalance is present & we have to rotate it
{
SplayNODE<WA> *k1 = k2->left; //create copy of temp's left & store it in k1
k2->left = k1->right; //make k1's right child as temp's left child
k1->right = k2; //make k1 as parent of temp node
k2 = k1; //assign k1 as new temp node (this is done because temp was passed by reference)
}
//==============================//==============================//
template <class WA>
void SplayTree<WA>::Zag(SplayNODE<WA> * & k2)
{
SplayNODE<WA> *k1 = k2->right; //create copy of temp's right child & store it in k1
k2->right = k1->left; //store k1's left child as temp's right child
k1->left = k2; //make k2 as left child of k1
k2 = k1; //assign k1 as temp node (due to reference of k2)
}
//==============================//==============================//
&这是我的**问题。首先，我只在搜索功能中展开。即如果找到具有特定密钥的节点，则展开。
我的搜索功能：
template <class WA>
SplayNODE<WA>* SplayTree<WA>::search(SplayNODE<WA> *temp, WA value) /////PVT DEFINITION
{
SplayNODE<WA> *to_searched = 0; //created new node pointer in which address of required node will be saved
if (temp!=0) //if arg. given temp node is not empty, then
{
if (temp->key==value) //if temp node has user-specified value then
{
to_searched = temp; //assign address of temp to to_searched node, which will be return @ the end
Do_Splay(temp); //perform splay at node which is found
}
if (to_searched==0) //if node is still empt then
{ to_searched = search(temp->left, value); } //recursive call to search in left sub-tree of temps
if (to_searched==0) //if node is still empt then
{ to_searched = search(temp->right, value); } //recursive call to search in right sub-tree of temps
}
return to_searched; //Finally return the searched node
}
//==============================//==============================//
如果节点是根的子节点，则以下工作正常。- Zig Single - Zag Single 但是，一旦我们关闭一个节点，问题就开始了。我无法成功执行任何这些事情。 之字形 - 之字形 - 之字形 - 之字形 - 之字形
这是样本树。（之字形案例）
20
/ \
10 25
/ \
5 15
当搜索到 5 时，答案应该是。
5
\
10
\
20
/ \
15 25
但我的输出变成：
20
/ \
15 25
无法弄清楚是什么问题。试运行这东西 100 次，但是。:( 任何和所有帮助表示赞赏。在此先感谢