zoukankan      html  css  js  c++  java
  • 浙大数据结构课后习题 练习三 7-4 List Leaves (25 分)

    Given a tree, you are supposed to list all the leaves in the order of top down, and left to right.

    Input Specification:

    Each input file contains one test case. For each case, the first line gives a positive integer N (≤) which is the total number of nodes in the tree -- and hence the nodes are numbered from 0 to N1. Then N lines follow, each corresponds to a node, and gives the indices of the left and right children of the node. If the child does not exist, a "-" will be put at the position. Any pair of children are separated by a space.

    Output Specification:

    For each test case, print in one line all the leaves' indices in the order of top down, and left to right. There must be exactly one space between any adjacent numbers, and no extra space at the end of the line.

    Sample Input:

    8
    1 -
    - -
    0 -
    2 7
    - -
    - -
    5 -
    4 6
    

    Sample Output:

    4 1 5


    #include <iostream>
    #include <queue>
    #include <vector>
    using namespace std;
    /**树结构体*/
    struct tn{
        int index;
        string lc;
        string rc;
    };
    /**层序遍历,返回一个vec*/
    vector<int> levelOrderTreserve(tn tree[],int index){
        vector<int> res;queue<tn> que;tn tmp;
        que.push(tree[index]);
        while(!que.empty()){
            tmp=que.front();
            que.pop();
            if(tmp.lc!="-") que.push(tree[(tmp.lc[0]-'0')]);
            if(tmp.rc!="-") que.push(tree[(tmp.rc[0]-'0')]);
            if(tmp.lc=="-"&&tmp.rc=="-") res.push_back(tmp.index);
        }
        return res;
    }
    int main()
    {
        /**input*/
        int T;
        cin>>T;
        tn tree[T];int judgeRoot[T];int rootPos;
        for(int i=0;i<T;i++) judgeRoot[i]=0;
        for(int i=0;i<T;i++){
            cin>>tree[i].lc>>tree[i].rc;tree[i].index=i;
            /**找出根节点*/
            if(tree[i].lc!="-") judgeRoot[tree[i].lc[0]-'0']++;
            if(tree[i].rc!="-") judgeRoot[tree[i].rc[0]-'0']++;
        }
        for(int i=0;i<T;i++) if(!judgeRoot[i]) rootPos=i;
        vector<int> res=levelOrderTreserve(tree,rootPos);
        for(int i=0;i<res.size();i++) {
            cout<<res[i];
            if(i!=res.size()-1) cout<<" ";
        }
        system("pause");
        return 0;
    }
  • 相关阅读:
    vue比较模板来跟新数据
    iframe自适应高度
    springmvc接口ios网络请求
    spring mvc实现接口参数统一更改
    spring mvc实现自定义注解
    基于redis集群实现的分布式锁,可用于秒杀,定时器。
    java使用javax.mail进行免费的邮件发送
    前端图片压缩上传(纯js的质量压缩,非长宽压缩)
    java项目中的路径获取,request
    阿里云(腾讯云类似)服务器控制台配置开放所有的端口
  • 原文地址:https://www.cnblogs.com/littlepage/p/11380800.html
Copyright © 2011-2022 走看看