## The order of a Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2738    Accepted Submission(s): 1454

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

``````#include <iostream>
#include <cstdio>

/*
* 二叉搜索树模板题
*/
using namespace std;

/*
* 定义节点
*/
typedef struct node {
int data;
node *lchild;
node *rchild;
} BSTNode, *BST;

bool isFirstTime = true;

/*
* 二叉树的插入操作.
*
*/
void insert(BST &T, int data) {

/*
* 如果当前二叉树为空
*/
if (T == NULL) {
//新建根节点
T = new BSTNode();

T->data = data;
T->lchild = NULL;
T->rchild = NULL;
return;
}

/*
* 如果插入的数值小于根节点的数值
*/
if (data < T->data) {
//插入左子树
insert(T->lchild, data);
} else {
//否则,插入右子树
insert(T->rchild, data);
}
}

/*
* 前序遍历
*/
void preOrder(BST &T) {

//如果当前的树不为空
if (T != NULL) {

//格式上的要求,和算法本身无关
if (isFirstTime == false) {
printf(" ");
}

isFirstTime = false;

//访问根节点
printf("%d", T->data);
//访问左子树
preOrder(T->lchild);
//访问右子树
preOrder(T->rchild);
}
}

int main() {

int n;

while (scanf("%d", &n) != EOF) {
BST T = NULL;

isFirstTime = true;
int i;
for (i = 0; i < n; ++i) {
int data;
scanf("%d", &data);
insert(T,data);
}

preOrder(T);

//不加会PE
printf("\n");
}

return 0;
}``````