<hdu

　　这里是杭电hdu上的链接：http://acm.hdu.edu.cn/showproblem.php?pid=3999

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:

　　1 3 4 2

Sample Output

　　1 3 2 4

　　解题思路：我的标题已经指出了这是水题，套用二叉搜索树的模板就行。另外在输出的时候注意空格即可，如果好几次没有ac的话可以参考下输出。

 1 #include <iostream>
 2 #include <cstdlib>
 3 using namespace std;
 4 
 5 int j;
 6 
 7 struct node {
 8     int val;
 9     node *lch;
10     node *rch;
11 };
12 
13 node *creat (node *p, int x) {
14     if (p == NULL) {
15         node *q = new node;
16         q->val = x;
17         q->lch = q->rch = NULL;
18         return q;
19     }
20     else {
21         if (x < p->val) p->lch = creat (p->lch,x);
22         else p->rch = creat (p->rch,x);
23         return p;
24     }
25 }
26 void pre_order (node *s) {
27     //int j = 0;如果写这里--->每一次进来都是0
28     if (s) {
29         if (j > 0)
30             cout << " ";
31         j++;
32         cout << s->val;
33         pre_order (s->lch);
34         pre_order (s->rch);
35     }
36 }
37 
38 int main() {
39     int n,x;
40     while (cin >> n) {
41         j = 0;
42         node *root = NULL;
43         for (int i = 0;i < n; ++i) {
44             cin >> x;
45             root = creat(root,x);
46         }
47         pre_order (root);
48         cout << endl;
49     }
50     return 0;
51 }

View Code

　　欢迎码友评论，一起成长。