这个程序的思路是自己输入数字,在输入的同时,已经帮你左右顺序排好了,即左子树的数字比右子树小,是个顺序二叉树,以输入0为结素,而后一中序遍历输出,但不知道为什么,在屏幕上打引的却是左子树最小的数字,而且一直输出,请看下面程序
#include <stdio.h>
#include <stdlib.h>
typedef struct no
{
int key;
struct no *left,*right;
}node,*PNOD;
void inster(PNOD *p,int k)
{
PNOD ppre,pre,temp;
ppre=*p;
printf("current number ppre % d",ppre->key);
if(ppre==NULL)
{
ppre=(node *)malloc(sizeof(node));
ppre->key=k;
ppre->left=NULL;
ppre->right=NULL;
*p=ppre;
return;
}
while(ppre)
{
if(k < ppre->key)
{
temp=ppre;
ppre=ppre->left;
printf("left\n");
}
else if(k == ppre->key)
{
printf("has ...\n");
return;
}
else if(k > ppre->key)
{
temp=ppre;
ppre=ppre->right;
printf("right\n");
}
/*printf("aaaaaaaaaaaaaaaaaa");*/
}
pre=(node *)malloc(sizeof(node));
pre->key=k;
pre->left=NULL;
pre->right=NULL;
#include <stdio.h>
#include <stdlib.h>
typedef struct no
{
int key;
struct no *left,*right;
}node,*PNOD;
void inster(PNOD *p,int k)
{
PNOD ppre,pre,temp;
ppre=*p;
printf("current number ppre % d",ppre->key);
if(ppre==NULL)
{
ppre=(node *)malloc(sizeof(node));
ppre->key=k;
ppre->left=NULL;
ppre->right=NULL;
*p=ppre;
return;
}
while(ppre)
{
if(k < ppre->key)
{
temp=ppre;
ppre=ppre->left;
printf("left\n");
}
else if(k == ppre->key)
{
printf("has ...\n");
return;
}
else if(k > ppre->key)
{
temp=ppre;
ppre=ppre->right;
printf("right\n");
}
/*printf("aaaaaaaaaaaaaaaaaa");*/
}
pre=(node *)malloc(sizeof(node));
pre->key=k;
pre->left=NULL;
pre->right=NULL;






