我正在向您发布使用我开发的AVL 树的代码。插入的方法,avlinsert
方法如下。我在纸上开发了这段代码,它没有经过测试,但我希望这能奏效。我要讨论的主要问题是节点的平衡因子首先看代码。通过这种方式,这个想法将变得清晰,我想问什么。所以这里是代码:
treeNode* avlinsert(treeNode* tree, int info)
{
treeNode* newNode=new treeNode;
newNode->setinfo(info);
newNode->setbalance(0);
treeNode* p,*q;
bool duplicate=false;
p=q=tree;
stack s; //I have made this stack already and now I am using it here.
//Now the the while loop block will check for duplicate nodes, this block prevents the duplicate insertion.
while (q!=NULL)
{
p=q;
if (info < p -> getinfo())
q=p->getleft();
else if (info>p->getinfo())
q=p->getright();
else
{
duplicate=true;
cout<<"Trying to insert duplicate";
break;
}
}//while loop ended.
//Now checking for duplicates.
if (duplicate)
return tree;
p=q=tree;
//Now below is the main block of while loop which calculates the balance factors of nodes and helps in inserting nodes at proper positions.
while (q!=NULL)
{
p=q;
if (info < p -> getinfo())
{
p->setbalance(p -> getbalance()+1);
s.push(p);//pushing into stack
q=p->getleft();
}
else if (info > p -> getinfo())
{
p->setbalance(p->getbalance()-1);
q=p->getright();
}
}//while loop ended
//Now the below code block will actually inserts nodes.
if (info < p -> getinfo())
p->setleft(newNode);
else if (info > p -> getinfo())
p->setright(newNode);
//After this insertion we need to check the balance factor of the nodesand perform the approprite rotations.
while (!s.isempty())
{
treeNode node;
node=s.pop();
int balance;
balance=node.getbalance();
if (balance==2)
{
s.Makeempty(); // We have found the node whoes balance factor is violating avl condition so we don't need other nodes in the stack, therefor we are making stack empty.
treeNode* k1,*k3;
k1=&node; //This is the node whoes balance factor is violating AVL condition.
k3=&(k1 -> getleft()); //Root of the left subtree of k1.
//Identifying the cases of insertion
if (info < k3 -> getinfo()) //This is the case of insertion in left subtree of left child of k1 here we need single right rotation.
root=Rightrotation(k1); //Here root is the private data member.
//Next case is the insertion in right subtree of left child of k1.
if (info > k3 ->getinfo())
root=LeftRightrotation(k1);
}//end of if statement.
}//while loop ended
这不是全部代码,但足以向您展示我正在尝试做的事情。您已经在这段代码中看到我在插入期间设置节点的平衡因子(第二个 while 循环块)。好的,这很好。但是在插入之后,我需要执行旋转。我也有旋转代码,但主要问题是当节点旋转时,我需要重置旋转代码中节点的平衡因子。这就是我的问题。我该怎么做?代码片段是什么?