问一个二叉树遍历的问题

[jianzhibeihang]

一颗二叉树，没个子节点都有一个指向起父结点的指针，让你在线性空间内遍历这颗树 ......

typedef struct node{
        int data;
        struct node *left;
        struct node *right;
        struct node *parent;
        }Node;

我做出来个死循环，谢谢大家指教！

[jianzhibeihang]

这是我的代码....

#include <stdio.h>
#include <stdlib.h>

typedef struct node{
        int data;
        struct node *left;
        struct node *right;
        struct node *next;
        }Node;

Node *FindLeft(Node *p)
{
     if(p == NULL)
          return NULL;
     if(p->left != NULL)
                p = p->left;
     return p;
}

void incursive(Node *root)
{
     Node *p = root;
     if(root == NULL)
             return;
     if(root->left == NULL && root->right == NULL)
                   printf("%d\n",root->data);
     
     if(p->left == NULL)
     {
                   printf("%d\n",p->data);
                   p = p->right;
                   p = FindLeft(p);      
     }
     else{
     //     printf("Hello1\n");
         p = FindLeft(p);
     }
 /*    
     if(p == NULL)
          printf("Fuck\n");
   */      
     while(p != NULL)
     {
         //    printf("Hello2!\n");
             printf("%d\n",p->data);
             if(p->right != NULL)
                         incursive(p->right);
             else
             {
                 printf("%d\n",p->data);
                 p = p->next;
                 if(p == root)
                      break;
             }
     }
}

int main()
{
    Node *p1,*p2,*p3,*p4,*p5,*root;
    root = (Node *)malloc(sizeof(Node));
    p1 = (Node *)malloc(sizeof(Node));
    p2 = (Node *)malloc(sizeof(Node));
    p3 = (Node *)malloc(sizeof(Node));
    p4 = (Node *)malloc(sizeof(Node));
    p5 = (Node *)malloc(sizeof(Node));
    
    
    root->data = 0;
    root->left = p1;
    root->right = p2;
    root->next = NULL;
    
    p1->data = 1;
    p1->left = NULL;
    p1->right = p3;
    p1->next = root;
    
    p2->data = 2;
    p2->left = NULL;
    p2->right = NULL;
    p2->next = root;
    
    p3->data = 3;
    p3->left = p4;
    p3->right = p5;
    p3->next = p1;
    
    p4->data = 4;
    p4->left = NULL;
    p4->right = NULL;
    p4->next = p3;
    
    p5->data = 5;
    p5->left = NULL;
    p5->right = NULL;
    p5->next = p3;
    
    incursive(root);
    
    free(root);
    free(p1);
    free(p2);
    free(p3);
    free(p4);
    free(p5);

    system("pause");
    return 0;
}

[budweiser]

楼主 我不能不说你的代码写的非常乱。。。

[budweiser]

楼主不妨自己调试下 看是哪里的问题、

[chj_2003]

[Quote=引用 10 楼 jianzhibeihang 的回复:]
引用 9 楼 chj_2003 的回复:

父节点指针是用来判断节点路径的


给个代码参考下吧？？
[/Quote]
if(p->next->left==p)左孩子；if(p->next->right==p)右孩子；推到根节点知道路径了

[pmars]

Node *FindLeft(Node *p)
{
     if(p == NULL)
          return NULL;
     if(p->left != NULL)
                p = p->left;
     return p;
}

void incursive(Node *root)
{
     Node *p = root;
     if(root == NULL)
             return;
     if(root->left == NULL && root->right == NULL)
                   printf("%d\n",root->data);
     
     if(p->left == NULL)
     {
                   printf("%d\n",p->data);
                   p = p->right;
                   p = FindLeft(p);      
     }
     else{
     //     printf("Hello1\n");
         p = FindLeft(p);  //第一次执行到这里的时候p被赋值为p1
     }
 /*    
     if(p == NULL)
          printf("Fuck\n");
   */      
     while(p != NULL)//之后执行这里
     {
         //    printf("Hello2!\n");
             printf("%d\n",p->data);
             if(p->right != NULL）//
                         incursive(p->right);//p1又右子树，所以执行，但是始终没有更改p的值，导致每次while循环都TRUE，死了就
             else
             {
                 printf("%d\n",p->data);
                 p = p->next;
                 if(p == root)
                      break;
             }
     }
}

[jianzhibeihang]

[Quote=引用 4 楼 pmars 的回复:]

C/C++ code

Node *FindLeft(Node *p)
{
if(p == NULL)
return NULL;
if(p->left != NULL)
p = p->left;
return p;
}

void incursive(Node *root)
{
Node *p = ro……
[/Quote]

怎么改呢？？？帮帮忙吧....

[chj_2003]

if(p == NULL)
printf("Fuck\n")；哥们，你这句很有意思。

[chj_2003]

#include <stdio.h>
#include <stdlib.h>

typedef struct node{
int data;
struct node *left;
struct node *right;
struct node *next;
}Node;



void incursive(Node *root)
{
Node *p = root;
if(root == NULL)
return;

printf("%d\n",p->data);
incursive(p->left);
incursive(p->right);
}









int main()
{
Node *p1,*p2,*p3,*p4,*p5,*root;
root = (Node *)malloc(sizeof(Node));
p1 = (Node *)malloc(sizeof(Node));
p2 = (Node *)malloc(sizeof(Node));


root->data = 0;
root->left = p1;
root->right = p2;
root->next = NULL;

p1->data = 1;
p1->left = NULL;
p1->right = p3;
p1->next = root;

p2->data = 2;
p2->left = NULL;
p2->right = NULL;
p2->next = root;

p3->data = 3;
p3->left = p4;
p3->right = p5;
p3->next = p1;

p4->data = 4;
p4->left = NULL;
p4->right = NULL;
p4->next = p3;

p5->data = 5;
p5->left = NULL;
p5->right = NULL;
p5->next = p3;

incursive(root);

free(root);
free(p1);
free(p2);
free(p3);
free(p4);
free(p5);

system("pause");
return 0;
}

[jianzhibeihang]

[Quote=引用 7 楼 chj_2003 的回复:]
#include <stdio.h>
#include <stdlib.h>

typedef struct node{
int data;
struct node *left;
struct node *right;
struct node *next;
}Node;



void incu……
[/Quote]
哥们你这里没用到指向父节点的指针啊？？？这不是这个题目的本意了啊

[chj_2003]

父节点指针是用来判断节点路径的

[jianzhibeihang]

[Quote=引用 9 楼 chj_2003 的回复:]

父节点指针是用来判断节点路径的
[/Quote]

给个代码参考下吧？？