设在一棵二叉搜索树的每个结点中,含有关键码key域和统计相同关键码

#include<stdio.h>
#include<stdlib.h>
typedef struct BTNode
{
	int data;
	int count;
	struct BTNode *lchild;
	struct BTNode *rchild;
}BTNode;
/*
int insert(BTNode *&T,int key)//递归方法解决
{
	if(T==NULL)
	{
		T=(BTNode*)malloc(sizeof(BTNode));
		T->data=key;
		T->count=1;
		T->lchild=T->rchild=NULL;
		return 1;
	}
	else if(key<T->data)
	insert(T->lchild,key);
	else if(key>T->data)
	insert(T->rchild,key);
	else
	{
		T->count++;
		return 0;
    }
}
*/
int insert(BTNode *&T,BTNode *&pr,int key)//非递归
{
	BTNode *p=T;
	while(p!=NULL)
	{
		if(p->data!=key)
		{
			pr=p;
			if(p->data>key)
			p=p->lchild;
			else if(p->data<key)
			p=p->rchild;
		}
		else
		{
			(p->count)++;  //此处如果找到，如果找不到两种情况
                                       //1、T为空树 2、T不为空，但没有key
			return 0;
		}
	}
	BTNode *s=(BTNode*)malloc(sizeof(BTNode));
	s->data=key;
	s->count=1;
	s->lchild=s->rchild=NULL;
	if(pr==NULL)
	T=s;
	else if(pr->data>key)
	pr->lchild=s;
	else
	pr->rchild=s;
	return 1;
}
void pre(BTNode *T)
{
	if(T!=NULL)
	{
		printf("\ndata=%d count=%d \n",T->data,T->count);
		pre(T->lchild);
		pre(T->rchild);
	}
}
int main()
{
	int a[]={1,3,2,1,2,3,2,7,6,5,5,6};
	BTNode *T=NULL,*pr=NULL;
	for(int i=0;i<12;i++)
	printf("%d ",insert(T,pr,a[i]));
	pre(T);
}