HDU 3999 The order of a Tree （排序二叉树的先序遍历）

题目

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

思路

给你一个序列，让你根据这个序列建立排序二叉树，然后问你一个序列，这个序列最后建立的排序二叉树和我给出的是一颗，且字典序最小。那么由排序二叉树的性质我可以知道，先序遍历的序列一定是字典序最小的。

Code

#include<vector>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
void check_max (int &a,int b) {a=max (a,b);}
void check_min (int &a,int b) {a=min (a,b);}
vector <int> v;
int n;
struct node {
	int data;
	node *leftchild,*rightchild;
    node () {
		leftchild=NULL;
		rightchild=NULL;
	}
};

void insert (node *&root,int data) {
	if (root==NULL) {
		root=new node ();
		root->data=data;
		return ;
	}
	else if (root->data==data) return ;
	else if (data>root->data) insert (root->rightchild,data);
	else if (root->rightchild,data) insert (root->leftchild,data);
	return ;
}

node * build () {
	node *root=NULL;
	for (auto it:v) {
		insert (root,it);
	}
	return root;	
}

void pre_travel (node *root,int flag) {
	if (root==NULL) return ;
	if (flag==1) printf ("%d",root->data);
	else printf (" %d",root->data);
	pre_travel (root->leftchild,++flag);
	pre_travel (root->rightchild,++flag);
	return ;
}

int main () {
    while (scanf ("%d",&n)!=EOF) {
		v.clear ();
		for (int i=1;i<=n;i++) {
			int x; scanf ("%d",&x);
			v.push_back (x);
		}
		node *root=build ();
		pre_travel (root,1);
		printf ("
");
	}
	return 0;
}