二叉排序树：HUD3999-The order of a Tree（二叉排序树字典序输出）

The order of a Tree

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2070 Accepted Submission(s): 1097

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

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解题心得：

1. 题意就是给你一串数，先生成一个二叉排序树，然后按照字典序输出就可以了。没有什么难点，只是在按照字典序输出的时候注意递归输出方式就可以了。

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#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+100;
int n,index;
struct node
{
    int num;
    node *lchild,*rchild;
} tree[maxn];

node* newnode(int num)
{
    tree[index].num = num;
    tree[index].lchild = NULL;
    tree[index].rchild = NULL;
    return &tree[index++];
}

void insertBST(node*& root,int num)
{
    if(root == NULL)
    {
        root = newnode(num);
        return ;
    }
    else if(root->num > num)
        insertBST(root->lchild,num);
    else
        insertBST(root->rchild,num);
}

void buildtree(node*& root)
{
    root = NULL;
    int num;
    for(int i=1; i<=n; i++)
    {
        scanf("%d",&num);
        insertBST(root,num);
    }
}

void print(node*& root,int num)
{
    if(root == NULL)
        return ;
    if(num == 2)
        printf(" ");
    printf("%d",root->num);//搜到一个输入一个
    print(root->lchild,2);//先向左方开始找
    print(root->rchild,2);
}

int main()
{
    while(scanf("%d",&n) != EOF)
    {
        node *p;
        index = 0;
        buildtree(p);//先构建二叉树
        print(p,1);//然后输出
        printf("\n");
    }
}