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:
    关于django模型里面的__str__和__unicode__
    js基础之javascript的存在形式和js代码块在页面中的存放位置和 CSS 对比
    css样式之边框和内外边距
    css样式之背景图片
    python学习笔记(2)--python3新特性
    linux命令每日一练习-rmdir mv
    代码规范
    抽象类和接口的区别
    单例模式及其并发问题
    使用url下载网络图片以及流介绍
  • 原文地址:https://www.cnblogs.com/littlepage/p/11380800.html
Copyright © 2011-2022 走看看