平衡二叉树的旋转问题

主要是要想好有哪些情况

，只有想清楚全部情况后，才能对BF进行判断



#include<stdio.h>
#include<stdlib.h>
typedef struct BiTre{
int data;
int bf;
struct BiTre *rchild,*lchild;
}BiTnood,*BiTree;
void R_Rotate(BiTree *p)
{
BiTree l = (*p)->lchild;
(*p)->lchild = l->rchild;
l->rchild = (*p);
(*p) = l;
}
void L_Rotate(BiTree *p)
{
BiTree l = (*p)->rchild;
(*p)->rchild = l->lchild;
l->lchild = (*p);
(*p)= l;
}
void LeftBalance(BiTree *T)//改变左子树的状态
{
BiTree L,Lr;
L = (*T)->lchild;
switch(L->bf)
{
case 1:
L->bf=0;
(*T)->bf = 0;
R_Rotate(T);
break;
case -1:
Lr = L->rchild;
switch(Lr->bf)
{
case 0:
L->bf=0;
(*T)->bf=0;
break;
case 1:
(*T)->bf = 1;
L->bf = 0;
break;
case -1:
(*T)->bf = 0;
L->bf = -1;
break;

}
Lr->bf = 0;
L_Rotate(&((*T)->lchild));
R_Rotate(T);

}
}
void ReftBalance(BiTree *T)//改变右子树的状态
{
BiTree R,Rl;
R = (*T)->rchild;
switch(R->bf)
{
case -1:
(*T)->bf = 0;
R->bf = 0;
R_Rotate(T);
break;
case 1:
Rl = R->lchild;
switch(Rl->bf)
{
case 0:
(*T)->bf=0;
R->bf=0;
break;
case 1:
(*T)->bf = 0;
R->bf = -1;
break;
case -1:
(*T)->bf = 1;
R->bf = 0;

}
Rl->bf=0;
R_Rotate(&((*T)->rchild));
L_Rotate(T);
}

}
int main()
{
return 0;
}