HDU - 3999 E - The order of a Tree

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

题意：给一个序列，构建二叉搜索树，输出树的前序遍历。

分析：用数组模拟二叉树，用数组记录二叉树前序遍历。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int s[100010],left[100010],right[100010];
int sum[100010];
int len1,len2,num;
void tree(int l,int k)
{ 
    if(s[l]>=k)    //左子树
    {
        if(left[l])
            tree(left[l],k);
        else
            left[l]=len2;
    }
    else           //右子树
    {
        if(right[l])
            tree(right[l],k);
        else
            right[l]=len2;
    }
}
void check(int l)   //树的前序遍历
{
    sum[num++]=s[l];   
    if(left[l])     //查找左子树
        check(left[l]);
    if(right[l])    //右子树
        check(right[l]);
}
int main()
{
    int n,k;
    while(~scanf("%d",&n))
    {
        memset(left,0,sizeof(left));
        memset(right,0,sizeof(right));
        len1=-1;
        len2=0;
        num=0;
        for(int i=0; i<n; i++)  //建树
        {
            scanf("%d",&k);
            if(len1==-1)        //根节点
                s[++len1]=k;
            else
            {
                s[++len2]=k;
                tree(len1,k);
            }
        }
        check(len1);  //前序遍历
        for(int i=0; i<num; i++)
        {
            if(i==0)
                printf("%d",sum[i]);
            else
                printf(" %d",sum[i]);
        }
        printf("\n");
    }
}