AVL树中的删除

正如您所知道的，在删除节点后应该如何平衡avl，我将说到点子上。首先，我正在考虑删除一个没有子节点的节点。

例如树：

10
/     
5      17
/      /   
2  9   12  20
           
3          50

假设deletevalue(12)

然后树应该在删除后：

10
/     
5      17
/          
2  9       20
           
3          50

现在，我们看到树在节点17处是平衡的，因为根据公式，它的平衡因子=高度(左子树[左树为空，因此-1])-高度(右子树)=-2

因此，我们通过检查树的大小写是右还是左来平衡树。

If BalanceFactor(17's right) = -1
perform SingleLeftRotation(17);
else if BalanceFactor(17's right) = -1
perform DoubleRightLeftRotation(17);

类似的情况是，如果17的平衡因子为2，即它保持高，则其相应的旋转。//对于bF(17)=2

If BalanceFactor(17's left) = 1
perform SingleLeftRotation(17);
else if BalanceFactor(17's left) = -1
perform DoubleLeftRightRotation(17);

平衡后，树应该变成这样：

10
/     
5      20
/      /   
2  9   17  50
           
3  

这是我设计的删除

从主要功能，我调用

bool deletevalue(WA value)
{
AvLNode<WA> *temp = search(root, value);    //calling search function to find node which has user-specified data & stored its address in temp pointer
if(temp!=0) //if temp node is not null then
{
if(temp->left==0 && temp->right==0) //if temp node don't have any children
{   deletewithNochild(root, value); }   //call to respective function
else if( (temp->left!=0 && temp->right==0) || (temp->left==0 && temp->right!=0) )   //if temp node has any 1 child, left or right
{   deletewithOneChild(temp);   }   //call to respective function
else if(temp->left!=0 && temp->right!=0)    //if temp node has 2 children
{   deletewith2Child(temp);     }   //call to respective function
return true;    //for prompting respective output message
}
else
return false;   //for prompting respective output message
}

由于我们所需的节点并没有子节点，所以下面的函数是envoked

void deletewithNochild(AvLNode<WA> *temp, WA value) //temp is node which is to be deleted
{
if(value == root->key)  //if temp is root node then
{
delete root;    //free memory of root node
root = 0;   //nullify root
}
else    //if temp is some other node 
{
if (value < temp->key)
{
deletewithNochild(temp->left, value);
}
else if (value > temp->key)
{
deletewithNochild(temp->right, value);
}
else if (value == temp->key)
{
AvLNode<WA> *father = findfather(temp, root);   //calling findfather func to find father of temp node & store its address in father node pointer
if(father->left==temp)  //if temp is left child of its father
{
delete temp;    //free memory of temp node
father->left=0; //nullify father's left
}
else if(father->right==temp)    //if temp is right child of its father
{
delete temp;    //free memory of temp node
father->right=0;//nullify father's right
}
return;
}
cout<<"nBalancing";
if ( balancefactor(temp) == 2)  //if temp is left higher, ie. temp's Balance Factor = 2, then
{
cout<<"t2 ";
if ( balancefactor(temp->left) == 1 ) //if temp's left node has Balance Factor 1 then
{
SingleRightRotation(temp);  //send temp node for rotation because temp is unbalance
}
else if ( balancefactor(temp->left) == -1 ) //if temp's left node has Balance Factor -1, then
{
DoubleLeftRightRotation(temp);  //send temp for double rotation because temp is unbalance
}
}
else if ( balancefactor(temp) == -2 )   //if temp is right higher, ie. temp's Balance Factor = -2, then
{
cout<<"t-2 ";
if ( balancefactor(temp->right) == -1 ) //if temp's left node has Balance Factor -1 then
{
SingleLeftRotation(temp);   //send temp node for rotation because temp is unbalance
}
else if ( balancefactor(temp->right) == 1 ) //if temp's right node has Balance Factor 1, then
{
DoubleRightLeftRotation(temp);  //send temp for double rotation because temp is unbalance
}
}
}
}

以下是节点的HEIGHT的两个实用函数&节点的平衡系数

int heightofnode(AvLNode<WA> *temp) const
{
return temp==NULL ? -1 : temp->height;
}

int balancefactor(AvLNode<WA> *temp) const
{
return ( heightofnode(temp->left) - heightofnode(temp->right) );
}

当我删除12时，我的输出变成(宽度第一遍历)-->[10][9][17]

请帮我，递归有问题吗？我又干流了&再次但无法理解。删除必须通过递归完成，否则平衡树将是一个更大的地狱。提前感谢您抽出时间。：-)