zoukankan      html  css  js  c++  java
  • hdu3999The order of a Tree (二叉平衡树(AVL))

    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
     
    Source
     
    #include<stdio.h>
    #include<malloc.h>
    typedef struct tree
    {
        struct tree *lchilde,*rchilde;
        int w;
    }tree,*Tree;
    void creaRoot(Tree &root)//创建根节点
    {
        root=(Tree)malloc(sizeof(tree));
        root->lchilde=root->rchilde=NULL;
    }
    void printTree(Tree root,int k)//遍历树并输出该节点的值
    {
        if(root==NULL) return ;//为空时反回
        if(k==0)
        printf("%d",root->w);
        else
        printf(" %d",root->w);
        printTree(root->lchilde,k+1);//先搜左
        printTree(root->rchilde,k+1);//后搜右
    }
    void InTree(Tree &node,int x)//建立二叉树,比当前节点小的放左边,否则放右边
    {
        Tree T;
        if(node==NULL)//为空时,就把该值放到该节点
        {
            T=(Tree)malloc(sizeof (tree));
            T->w=x;  T->lchilde=T->rchilde=NULL;//把左右孩子置为空
            node=T;
        }
        else//不为空,继续搜
        {
            if(x<=node->w)
            InTree(node->lchilde,x);
            else
            InTree(node->rchilde,x);
        }
    }
    int main()
    {
        Tree root;
        int n,i,x;
        while(scanf("%d",&n)>0)
        {
            if(n==0)continue;
            creaRoot(root);
            scanf("%d",&x);
            root->w=x;
            for(i=2;i<=n;i++)
            {
                scanf("%d",&x);
                InTree(root,x);
            }
            printTree(root,0);
            printf("
    ");
        }
    }
    



  • 相关阅读:
    01-SCOTT-SQL建表语句
    面试小题
    Swagger2使用API
    Elasticsearch单机安装_集群搭建_索引基本操作_Head插件安装与基本操作_ik分词器配置_logstash实现数据同步_教程
    45个非常有用的 Oracle 查询语句小结
    oracle获取定时时间段
    彻底理解webservice SOAP WSDL
    MySQL 1045登录失败
    quartz定时任务时间设置
    map 遍历的四种方式
  • 原文地址:https://www.cnblogs.com/riskyer/p/3268658.html
Copyright © 2011-2022 走看看