HDU 3999----The order of a Tree(二叉树的前序遍历)

本文链接：https://blog.youkuaiyun.com/W_904038290/article/details/76946417

本文介绍如何根据给定序列构建二叉搜索树，并寻找能够生成相同形状二叉搜索树的最小字典序序列。通过分析二叉搜索树的性质，文章给出了一种简单有效的解决方案——输出构建过程中的前序遍历序列。

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题目大意：给你一个序列，让你用这个序列建立一个二叉搜索树。然后在让你在输出一个序列，让这个序列可以构成和上面一样的二叉搜索树，且序列的字典序最小。

根据二叉搜索树的定义，左子树小右子树大，所以直接输出先序遍历就行了！！！也就是用题目给的序列建好二叉搜索树，然后输出他的前序遍历（最近刚开始学数据结构，为毛数据结构都是指针啊！！！最怕的就是指针，因为C语言没好好学他，没办法只能克服了），下面的代码大家共勉，有什么问题还请多多指教，我学的也不好，共同进步！

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

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

typedef struct Tnode
{
    int v;
    struct Tnode *left,*right;
}Node;//建立结点

Node *Buildtree(Node *root,int n)  //建树
{
    if(root==NULL)
    {
        root=(Node*)malloc(sizeof(Node));
        root->v=n;
        root->left=root -> right =NULL;
        return root;
    }
    if(root ->v>n)
        root -> left= Buildtree(root -> left,n);
    else
        root -> right=Buildtree(root -> right,n);
    return root;              //每次返回头指针，才能进行下一次的比较
}
void DLR(Node *root,int n) //前序遍历
{
    if(n!=1)
        printf(" %d",root -> v);
    else
        printf("%d",root->v);
    if(root->left!=NULL)
        DLR(root->left,2);
    if(root->right!=NULL)
        DLR(root->right,2);
}
int main()
{
    int n,m,i;
    Node *root=NULL;
    while(~scanf("%d",&n))
    {
        for(i=0;i<n;i++)
        {
            scanf("%d",&m);
            root=Buildtree(root,m);
        }
        DLR(root,1);
        printf("\n");
    }
    return 0;
}