标题：AVL树的基本操作例程（1）

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;}