The order of a Tree

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1361 Accepted Submission(s): 695

Problem Description

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.

Input

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.

Output

One line with n intergers, which are the order of a tree that generate the same tree with the least lexicographic.

Sample Input

4 1 3 4 2

Sample Output

1 3 2 4

题解：题目意思即，给你一个插入数列，形成一棵二叉树，你给出字典序最小的插入方法建相同的一棵树出来。说白了，就是求先序序列。

代码：

 1 #include <fstream>
 2 #include <iostream>
 3 
 4 using namespace std;
 5 
 6 typedef struct Node{
 7     Node *lch,*rch,*nex;
 8     int x;
 9     Node(int x){
10         this->x=x;
11         lch=NULL;
12         rch=NULL;
13     }
14 }inode;
15 
16 int n,tn;
17 inode *head;
18 
19 void insert(int t);
20 void preOrder(inode *p);
21 
22 int main()
23 {
24     //freopen("D:\input.in","r",stdin);
25     //freopen("D:\output.out","w",stdout);
26     int t;
27     while(~scanf("%d",&n)){
28         scanf("%d",&t);
29         head=new inode(t);
30         for(int i=1;i<n;i++){
31             scanf("%d",&t);
32             insert(t);
33         }
34         tn=0;
35         preOrder(head);
36     }
37     return 0;
38 }
39 void insert(int t){
40     inode *p=head,*s=new inode(t);
41     while(p!=NULL){
42         if(t<p->x)
43             if(p->lch!=NULL)    p=p->lch;
44             else{
45                 p->lch=s;
46                 break;
47             }
48         else
49             if(p->rch!=NULL)    p=p->rch;
50             else{
51                 p->rch=s;
52                 break;
53             }
54     }
55 }
56 void preOrder(inode *p){
57     if(p!=NULL){
58         printf("%d",p->x);
59         if(++tn<n)  printf(" ");
60         else    printf("
");
61         preOrder(p->lch);
62         preOrder(p->rch);
63         delete p;
64     }
65 }