从头到尾打印链表

题目：输入一个链表的头节点，从尾到头反过来打印出每个节点的值。链表的定义如下：

typedef int DataType;

typedef struct Node
{
DataType data;
struct Node *next;
}Node, *pList, *pNode;

【解题思路】（递归）

先递归出后面的节点，在输出节点本身。不足：当链表比较长时，会使函数调用层级比较深而导致函数调用栈溢出。

【代码】

void ReversePrint(pList phead)
{
if (phead != NULL)
{
if (phead->next)
{
ReversePrint(phead->next);
}
printf("%d   ", phead->data);
}
}

【程序】

#define _CRT_SECURE_NO_WARNINGS 1
# include <stdio.h>
# include <stdlib.h>
# include <assert.h>

typedef int DataType;

typedef struct Node
{
DataType data;
struct Node *next;
}Node, *pList, *pNode;

void InitLinkList(pList *pplist)
{
assert(pplist);
*pplist = NULL;
}

void PushBack(pList *pplist, DataType x)
{
pNode cur = *pplist;
pNode pnode = NULL;
assert(pplist);
pnode = (pNode)malloc(sizeof(Node));
if (pnode == NULL)
{
perror("PushBack::malloc");
exit(EXIT_FAILURE);
}
pnode->data = x;
pnode->next = NULL;
if (cur == NULL)
{
*pplist = pnode;
}
else
{
while (cur->next != NULL)
{
cur = cur->next;
}
cur->next = pnode;
}
}

void Display(pList plist)
{
pNode cur = plist;
while (cur)
{
printf("%d-->", cur->data);
cur = cur->next;
}
printf("over\n");
}

void Destroy(pList *pplist)
{
pNode cur = *pplist;
a
4000
ssert(pplist);
while (cur)
{
pNode del = cur;
cur = cur->next;
free(del);
}
*pplist = NULL;
}

void ReversePrint(pList phead)
{
if (phead != NULL)
{
if (phead->next)
{
ReversePrint(phead->next);
}
printf("%d   ", phead->data);
}
}

void test1()
{
pList plist;
pNode ret = NULL;
InitLinkList(&plist);
PushBack(&plist, 1);
PushBack(&plist, 2);
PushBack(&plist, 3);
PushBack(&plist, 4);
PushBack(&plist, 5);
PushBack(&plist, 6);
Display(plist);
ReversePrint(plist);
Destroy(&plist);
}

int main()
{
test1();
system("pause");
return 0;
}

运行结果：



【附加】（栈方法）

每经过一个节点的时候把节点放到一个栈中，遍历完节点后再从栈顶开始逐个输出节点的值。

【代码】

void ReversePrint(pList phead)
{
std::stack<pList>nodes;
pNode pnode = phead;
while (pnode != NULL)
{
nodes.push(pnode);
pnode = pnode->next;
}
while (!nodes.empty())
{
pnode = nodes.top();
printf("%d\t", pnode->data);
nodes.pop();
}
}