浙大数据结构课后习题 练习三 7-4 List Leaves (25 分)

浙大数据结构课后习题 练习三 7-4 List Leaves (25 分)
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大家好,我是架构君,一个会写代码吟诗的架构师。今天说一说浙大数据结构课后习题 练习三 7-4 List Leaves (25 分),希望能够帮助大家进步!!!

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;
}

 

转载于:https://www.cnblogs.com/littlepage/p/11380800.html

本文来源weixin_30595035,由架构君转载发布,观点不代表Java架构师必看的立场,转载请标明来源出处:https://javajgs.com/archives/29435

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