The order of a Tree

Problem Description

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

二叉排序树如何建立和遍历是关键。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int ld[100010],rd[100010],a,num,root,i;
void build(int root,int al)
{
    if(al>root)
    {
        if(rd[root]==-1)
        {
            rd[root]=al;
            //cout<<"al:"<<al<<" r root:"<<root<<endl;
        }
        else build(rd[root],al);
    }
    else
    {
        if(ld[root]==-1)
        {
            ld[root]=al;
            //cout<<"al:"<<al<<" l root:"<<root<<endl;
        }
        else build(ld[root],al);
    }
}

void solve(int root)
{
    if(ld[root]!=-1)
    {
        cout<<" "<<ld[root];
        solve(ld[root]);
    }
    if(rd[root]!=-1)
    {
        cout<<" "<<rd[root];
        solve(rd[root]);
    }
    else return;
}

int main()
{
    while(~scanf("%d",&num))
    {
        memset(ld,-1,sizeof(ld));
        memset(rd,-1,sizeof(rd));
        for(i=1;i<=num;i++)
        {
            scanf("%d",&a);
            if(i==1){root=a;}
            else build(root,a);
        }
        cout<<root;
        solve(root);
        cout<<endl;
    }
    return 0;
}