二叉树遍历中序,该怎么处理

2012-10-27

二叉树遍历中序C/C++ code#include stdio.h#include stdlib.h# define ERROR -2#define OK 1# define

二叉树遍历中序

C/C++ code

#include <stdio.h>#include <stdlib.h># define ERROR -2#define OK 1# define SIZE 100#define SIZE_IN 10typedef int Status ;typedef struct{    int * base;    int * top;    int sizebase;}Stack;typedef struct Node{    char data;    struct Node * lchild;    struct Node * rchild;}BTNode;Status InitStack(Stack &S){    S.base=(int *)malloc(SIZE*sizeof(int));    if(!S.base)    return ERROR;    S.top=S.base=0;    S.sizebase = SIZE;    return OK;}int EmptyStack(Stack S){    if(S.top == S.base)    return 1;    else return 0;}int Push(Stack &S,BTNode B){    if(S.top -S.base>=S.sizebase)    {        S.base = (int *)realloc(S.base,S.sizebase+(SIZE+SIZE_IN)*sizeof(int));        if(!S.base)        return 0;        (S).top=(S.sizebase)+(S.base);        S.sizebase= S.sizebase+SIZE_IN;    }    S.top++;    *S.top = (int)B.data;}int Pop(Stack &S,BTNode B){    if(S.top - S.base == 0)    return 0;    S.top --;    B.data=(char)*S.top;    return 1;}Status Create_BTNode(BTNode *T){    char ch;    scanf("%c",&ch);    if(ch == '#')        T == NULL;    else {        T = (BTNode *)malloc(sizeof(BTNode));        //分配内存        T->data = ch;//把ch付给data；        if(!T)        return ERROR;        Create_BTNode(T->lchild);//利用递归分别建立左右子树        Create_BTNode(T->rchild);        }        return OK;}void MidOrderTraverse(BTNode * B){    Stack S;    InitStack(S);    while(B || !EmptyStack(S))    {        if(B)        {            Push(S,B);            B=B->lchild;//86        }else{            Pop(S,B);            printf("%c",B->data);            B=B->rchild;//89        }    }}int main(){    return  0;}

D:\数据结构\BinaryTree\main.cpp:86: error: conversion from 'BTNode*' to non-scalar type 'BTNode' requested
D:\数据结构\BinaryTree\main.cpp:89: error: conversion from 'BTNode*' to non-scalar type 'BTNode' requested
这个是怎么回事，我看不出错误，谢谢大神，还有.和->有什么区别呀，还有我使用的形参是&的话，使用->就会说error，说木有指针，但是用.就对了，帮我看看吧大神，谢谢啦

[解决办法]
根据你代码改后如下：

C/C++ code

#include <stdio.h>#include <stdlib.h># define ERROR -2# define OK 1# define SIZE 100# define SIZE_IN 10typedef int Status ;typedef struct Node{    char data;    struct Node * lchild;    struct Node * rchild;}BTNode;typedef struct{    BTNode ** base;    BTNode ** top;    int sizebase;}Stack;Status InitStack(Stack &S){    S.base=(BTNode **)malloc(SIZE*sizeof(BTNode*));    if(!S.base)        return ERROR;    S.top=S.base;    S.sizebase = SIZE;    return OK;}int EmptyStack(Stack S){    if(S.top == S.base)        return 1;    else return 0;}int Push(Stack &S,BTNode *B){    if(S.top -S.base>=S.sizebase)    {        S.base = (BTNode **)realloc(S.base,S.sizebase+(SIZE+SIZE_IN)*sizeof(BTNode *));        if(!S.base)            return 0;        (S).top=(S.sizebase)+(S.base);        S.sizebase= S.sizebase+SIZE_IN;    }    *S.top = B;    S.top++;    return 1;}int Pop(Stack &S,BTNode **B){    if(S.top - S.base == 0)        return 0;    S.top --;    *B=*S.top;    return 1;}//建立二叉树，二叉树最好全部用cStatus Create_BTNode(BTNode **T){    char ch;    scanf("%c",&ch);    if(ch == '#')        *T = NULL;    else {        *T = (BTNode *)malloc(sizeof(BTNode));        //分配内存        (*T)->data = ch;//把ch付给data；        if(!T)            return ERROR;        Create_BTNode(&((*T)->lchild));//利用递归分别建立左右子树        Create_BTNode(&((*T)->rchild));    }    return OK;}//二叉树遍历中序输出，非递归void MidOrderTraverse(BTNode * B)//因为是*B，因此入栈是*B而不是B，B是个地址，因此加上*B{    Stack S;    InitStack(S);    while(B || !EmptyStack(S))    {        if(B)        {            Push(S,B);            B=B->lchild;        }else{            Pop(S,&B);            printf("%c",B->data);            B=B->rchild;        }    }}int main(){    BTNode * root;    Create_BTNode(&root);    MidOrderTraverse(root);    system("pause");    return  0;} 
 
[解决办法]
C/C++ code#include <stdlib.h>#include <stdio.h>#include <string.h>#define NULL 0typedef struct tree{    char Data;    struct tree *left,*right;}*Bitree;Bitree create(Bitree T){    char ch;    scanf("%c",&ch);    if(ch=='#')//输入#号表示空    {        T=NULL;    }    else    {    if(!(T=(Bitree )malloc(sizeof(Bitree))))    {        printf("错误!\n");    }    T->Data=ch;    T->left=create(T->left);    T->right=create(T->right);}return T;}void xianxu(Bitree T){    if(T)    {        printf("%c",T->Data);        xianxu(T->left);        xianxu(T->right);    }}void zhongxu(Bitree T){    if(T)    {        zhongxu(T->left);        printf("%c",T->Data);        zhongxu(T->right);    }}void houxu(Bitree T){    if(T)    {        houxu(T->left);        houxu(T->right);        printf("%c",T->Data);    }}void main(){Bitree T;T=create(T);xianxu(T);printf("\n");zhongxu(T);printf("\n");houxu(T);printf("\n");}非递归先中后的遍历算法C/C++ code//1.先序遍历非递归算法#define maxsize 100typedef struct{    Bitree Elem[maxsize];    int top;}SqStack;void PreOrderUnrec(Bitree t){    SqStack s;    StackInit(s);    p=t;    while (p!=null || !StackEmpty(s))    {        while (p!=null)             //遍历左子树        {            visite(p->data);            push(s,p);            p=p->lchild;               }//endwhile                if (!StackEmpty(s))         //通过下一次循环中的内嵌while实现右子树遍历        {            p=pop(s);            p=p->rchild;                }//endif     }//endwhile }//PreOrderUnrec//2.中序遍历非递归算法#define maxsize 100typedef struct{    Bitree Elem[maxsize];    int top;}SqStack;void InOrderUnrec(Bitree t){    SqStack s;    StackInit(s);    p=t;    while (p!=null || !StackEmpty(s))    {        while (p!=null)             //遍历左子树        {            push(s,p);            p=p->lchild;        }//endwhile        if (!StackEmpty(s))        {            p=pop(s);            visite(p->data);        //访问根结点            p=p->rchild;            //通过下一次循环实现右子树遍历        }//endif       }//endwhile}//InOrderUnrec//3.后序遍历非递归算法#define maxsize 100typedef enum{L,R} tagtype;typedef struct {    Bitree ptr;    tagtype tag;}stacknode;typedef struct{    stacknode Elem[maxsize];    int top;}SqStack;void PostOrderUnrec(Bitree t){    SqStack s;    stacknode x;    StackInit(s);    p=t;   do     {        while (p!=null)        //遍历左子树        {            x.ptr = p;             x.tag = L;         //标记为左子树            push(s,x);            p=p->lchild;        }    while (!StackEmpty(s) && s.Elem[s.top].tag==R)          {            x = pop(s);            p = x.ptr;            visite(p->data);   //tag为R，表示右子树访问完毕，故访问根结点               }        if (!StackEmpty(s))        {            s.Elem[s.top].tag =R;     //遍历右子树            p=s.Elem[s.top].ptr->rchild;                }        }while (!StackEmpty(s));}//PostOrderUnrec

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二叉树遍历中序,该怎么处理