摘要(1)AVL树是带有平衡条件的二叉平衡树。它要求左子树和右子树的高度之差不能超过1;
(2)当插入一个新的节点后,可能会造成新的不平衡;因此要通过一定的操作来修正;我们将它称为旋转;
(3)不同的不平衡情况对应不同的旋转:对于一个平衡的节点有四种情况:对它的左儿子的左子树,左儿子右子树,右儿子左子树,右儿子的右子树;这对应了四种旋转;
(4)单旋转(左单旋,右单旋),双旋转(左双旋,右双旋);
如图所示:
代码:对双旋转提供了两种思路:调用两次但旋转/直接实现;
Position SingleRotateLeft(Position T)
{
static int amount = 0;
Position k1,k2;
k1 = T;
k2 = T->Left;
k1->Left = k2->Right;
k2->Right = k1;
amount ++ ;
printf("the times of splayingL is:%d\n",amount);
return k2;
}
Position SingleRotateRight(Position T)
{
static int amount = 0;
Position k1,k2;
k1 = T;
k2 = T->Right;
k1->Right = k2->Left;
k2->Left = k1;
amount ++;
printf("the times of splayingR is:%d\n",amount);
return k2;
}
Position DoubleRotateLeft(Position k3)
{
k3->Left = SingleRotateRight(k3->Left);
k3 = SingleRotateLeft(k3);
return k3;
}
Position DoubleRotateRight(Position k3)
{
k3->Right = SingleRotateLeft(k3->Right);
k3 = SingleRotateRight(k3);
return k3;
}
Position RotateOneRight(Position k1)
{
static int amount4 = 0;
Position k2 = k1->Right;
Position k3 = k2->Right;
k1->Right = k2->Left;
k2->Left = k1;
k2->Right = k3->Left;
k3->Left = k2;
amount4++;
printf("the times of splaying3 is:%d\n",amount4);
return k3;
}
Position RotateOneLeft(Position k1)
{
static int amount3 = 0;
Position k2 = k1->Left;
Position k3 = k2->Left;
k1->Left = k2->Right;
k2->Right = k1;
k2->Left = k3->Right;
k3->Right = k2;
amount3++;
printf("the times of splaying3 is:%d\n",amount3);
return k3;
}