Hdu - 3999- The order of a Tree

先上题目

　　

The order of a Tree

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 830 Accepted Submission(s): 448


[align=left]Problem Description[/align]
As we know,the shape of a binary search tree is greatly related to the order of keys we insert. To be precisely:
1. insert a key k to a empty tree, then the tree become a tree with
only one node;
2. insert a key k to a nonempty tree, if k is less than the root ,insert
it to the left sub-tree;else insert k to the right sub-tree.
We
call the order of keys we insert “the order of a tree”,your task
is,given a oder of a tree, find the order of a tree with the least
lexicographic order that generate the same tree.Two trees are the same
if and only if they have the same shape.

[align=left]Input[/align]
There
are multiple test cases in an input file. The first line of each
testcase is an integer n(n <= 100,000),represent the number of
nodes.The second line has n intergers,k1 to kn,represent the order of a
tree.To make if more simple, k1 to kn is a sequence of 1 to n.

[align=left]Output[/align]
One line with n intergers, which are the order of a tree that generate the same tree with the least lexicographic.

[align=left]Sample Input[/align]

4

1 3 4 2

[align=left]Sample Output[/align]

1 3 2 4

　　级别是水题，可是还是wa了一次，还是明知道会wa的还是交了上去，题目的意思就是给你一棵二叉排序树的输入顺序，求出这棵树的先序遍历。
　　这一题用指针来做的话很简单，但是我一开始用数组，而且知道2^100000一定会爆，但是还是用了数组来做= =，好吧我手贱。然后改成指针来做以后就过了，用时125MS。随后经人提醒用静态链表做，于是花了大概10分钟写出来了一个静态链表版的(好像更简单了= =)，交上去1y，用时95MS，看来静态链表是个好东西啊，而且听说java没有指针，只可以用静态链表= =。

上代码

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <stack>
#define MAX 100000+10
using namespace std;

typedef struct node
{
int d;
struct node* l,*r;
}node;

node *p;

void insert(node *f)
{
node *root;
root=p;
if(p==NULL) p=f;
else
{
while(1)
{
if(root->d > f->d)
{if(root->l)root=root->l;else {root->l=f;return;}}
else
{if(root->r)root=root->r;else {root->r=f;return;}}
}
}
}

void load()
{
int count;
node* f;
stack<node*> s;
f=p;
count=0;
s.push(f);
while(!s.empty())
{
f=s.top();
if(count++) printf(" ");
printf("%d",f->d);
s.pop();
if(f->r) s.push(f->r);
if(f->l) s.push(f->l);
free(f);
}
}

int main()
{
int i,n,e,count;
while(scanf("%d",&n)!=EOF)
{
node* f;
p=NULL;
for(i=0;i<n;i++)
{
scanf("%d",&e);
f=(node*)malloc(sizeof(node));
f->d=e;
f->l=f->r=NULL;
insert(f);
}
count=0;
load();
printf("\n");
}
return 0;
}

指针版