问一个关于建立二叉树的问题
程序是挺长的,呵呵!麻烦帮帮忙 我想问:
假如输入five,six,two三个单词
问tree函数每次返回的P是指向那里的?
define maxword 20
mian()
{struct tnode *root,*tree();
char *t;
root=null;
while((t=getword())!=NULL)
root=tree(root,t);
treeprint(root);
}
char *getword()
{char *p;
if((p=malloc(maxword+1))==NULL)
{printf( "out of memory\n ");
return(NULL);}
scanf( "%maxwords ",p);
if(strlen(p)==0)return(NULL);
return(p);
}
struct tnode *tree(p,w)
struct tnode *p;
char *w;
{struct tnode *talloc();
int cound;
if(p==NULL)
{p==talloc();
p-> word=w;
p-> count=1;
p-> left=p-> right=NULL;}
else if((cound=strcmp(w,p-> word))==0)
p-> count++;
else if(cound <0)
p-> left=tree(p-> left,w);
else
p-> right=tree(p-> right,w);
return(p);}
treeprint(p)
struct tnode *p;
{if(p!=NULL)
{treeprint(p-> left);
print( "%4d%s\n ",p-> count,p-> word);
treeprint(p-> ringht);}
}
struct tnode *talloc()
{return((struct tnode *)malloc(sizeof(struct tnode)));
}
[解决办法]
楼主也看的吉林大学那个教程吧
这道题我也做了,不过没调通,请前辈帮忙看看。
//生成一棵二叉树,比较单词出现的次数
//c++ java Fortran basic Foxbase
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAXWORD 20
struct tnode{
char *word; //指向一个单词
int count; //访问次数
struct tnode *left; //左孩子
struct tnode *right; //右孩子
};
char *GetWord()
{
char *p;
if((p=(char*)malloc(MAXWORD+1))==NULL) {printf( "Out of memory!\n ");return(NULL);}
scanf( "%s ",p);
if(strlen(p)==0) return(NULL);
return p;
}
struct tnode *tree(struct tnode *pp,char *w)
{
struct tnode *talloc();
int cond;
if(pp==NULL)
{
pp=talloc();
pp-> word=w;
pp-> count=1;
pp-> left=pp-> right=NULL;
}
else if((cond=strcmp(w,pp-> word))==0)
pp-> count++;
else if(cond <0)
pp-> left=tree(pp-> left,w);
else
pp-> right=tree(pp-> right,w);
return(pp);
}
void treeprint(struct tnode *point)
{
if(point!=NULL)
{
treeprint(point-> left);
printf( "%4d%s\n ",point-> count,point-> word);
treeprint(point-> right);
}
}
struct tnode *talloc()
{
return((struct tnode*)malloc(sizeof(struct tnode)));
}
int main()
{
struct tnode *root;
char *t;
root=NULL;
while((t=GetWord())!= "EOF ")
root=tree(root,t);
treeprint(root);
return 0;
}
[解决办法]
应指向这个树的根结点
[解决办法]
cound=strcmp(w,p-> word)
左子树,右子树,都是由 cound 决定的,
root-> word=five
root-> right-> word=six(six> five cound> 0)
root-> right-> right-> word=two(two> five cound> 0 在five的右子树中
two> six cound> 0 )
[解决办法]
为什么不用递归呢