The order of a Tree

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题目：

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：

输入一串数字，看是否能找到一串数字构成的二叉搜索树和输入数字构成的二叉搜索树相同，同时保证输出的字典序最小。

思路：

二叉搜索树有一个性质，先序遍历输出即为从小到大排好的序列。

根据输入的数字构建二叉搜索树，先序遍历输出即可。

代码：

#include<stdio.h>
#include<stdlib.h>
int k;
typedef struct note
{
    int data;
    struct note *left;
    struct note *right;
} Note;
typedef struct
{
    Note *root;
} Tree;
void insert(Tree *tree,int value)
{
    Note *note = malloc(sizeof(Note));
    note ->data =value;
    note ->left =NULL;
    note ->right=NULL;
    if(tree->root==NULL)
    {
        tree -> root=note;
    }
    else
    {
        Note *temp= tree ->root;
        while(temp != NULL)
        {
            if(value<temp -> data)
            {
                if(temp->left==NULL)
                {
                    temp->left=note;
                    return;
                }
                else
                {
                    temp=temp ->left;
                }
            }
            if(value>temp->data)
            {
                if(temp->right==NULL)
                {
                    temp->right=note;
                    return;
                }
                else
                {
                    temp=temp->right;
                }
            }
        }
    }
}
void perorder(Note *note)
{
    if(note !=NULL)
    {   if(k==0)
        printf("%d",note -> data);
        else
        printf(" %d",note -> data);
        k++;
        perorder(note -> left);
        perorder(note -> right);
    }
}
int main()
{
    int k;
    int b[110000];
    while(~scanf("%d",&k))
    {
        Tree tree;
        tree.root=NULL;
        int a,i;
        for(i=0; i<k; i++)
        {
            scanf("%d",&a);
            insert(&tree,a);
        }
        k=0;
        perorder(tree.root);
        printf("\n");
    }
    return 0;
}

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