Question

如果您知道在删除节点后应该如何平衡avl，我会指出。为了开始，我考虑删除一个没有孩子的节点。

例如树：

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

让我们说deletevalue（12）;

然后Tree应该在删除之后：

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

现在，我们看到树在节点17处是平衡的，因为通过公式，它的平衡因子=高度（左子树[左侧树是空的所以-1]） - 高度（右子树）= -2

所以我们通过检查它的右右情况还是左右情况来平衡树。

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

如果17的平衡因子是2，即它保持高，则其相应的旋转，情况类似。 //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，树应该变为：

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

这是我设计的删除。

从主要功能，我打电话

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
}

因为我们的必需节点没有子节点，所以调用了以下函数。

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
            }
        }

    }
}

以下是节点和高度的两个实用功能。节点的平衡因子

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) );
}

我的输出，当我删除12时 （广度优先交易者） - ＆gt;＆gt; [10] [9] [17]

请帮帮我，递归有什么问题吗？我再次干涸了再次但无法理解。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。： - ）

Answer 1

为什么deletewithNochild（）调用任何其他delete *方法？如果调用了deletewithNochild，则表示您处于要删除的节点。只需删除它，向上移动到它的父级，然后检查父级平衡因子并根据需要旋转。为每个节点的父节点重复重新平衡，直到到达根目录。

如果它有帮助，我已经实现了AVL tree in Java，如果你想要一个参考。

在AVL树中删除

1 个答案: