pat03-树2. List Leaves (25)

03-树2. List Leaves (25)

时间限制
400 ms
内存限制
65536 kB
代码长度限制
8000 B
判题程序
Standard
作者
CHEN, Yue

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 (<=10) which is the total number of nodes in the tree -- and hence the nodes are numbered from 0 to N-1. 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

提交代码

 

  1 #include<cstdio>
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<cstring>
  5 #include<stack>
  6 #include<queue>
  7 #include<vector>
  8 using namespace std;
  9 struct node{
 10     int l,r,f;
 11     node(){
 12         l=r=f=-1;
 13     }
 14 };
 15 node tree[10];
 16 /*void prefind(int h){//检验程序
 17     cout<<h<<endl;
 18     if(tree[h].l>=0)
 19         prefind(tree[h].l);
 20     if(tree[h].r>=0)
 21         prefind(tree[h].r);
 22 }*/
 23 int main(){
 24     //freopen("D:\\INPUT.txt","r",stdin);
 25     //freopen("D:\\OUTPUT.txt","w",stdout);
 26     int n;
 27     string s;
 28     scanf("%d",&n);
 29 
 30     //cout<<n<<endl;
 31 
 32     int i,num,j;
 33 
 34     /*for(i=0;i<n;i++){
 35         cout<<"f: "<<tree[i].f<<endl;
 36         cout<<"l: "<<tree[i].l<<endl;
 37         cout<<"r: "<<tree[i].r<<endl;
 38     }*/
 39 
 40 
 41    for(i=0;i<n;i++){
 42         cin>>s;
 43         if(s!="-"){
 44            num=0;
 45            j=0;
 46            while(j<s.length()){
 47               num*=10;
 48               num+=s[j++]-'0';
 49            }
 50            tree[i].l=num;
 51            tree[num].f=i;
 52         }
 53         cin>>s;
 54         if(s!="-"){
 55            num=0;
 56            j=0;
 57            while(j<s.length()){
 58               num*=10;
 59               num+=s[j++]-'0';
 60            }
 61            tree[i].r=num;
 62            tree[num].f=i;
 63         }
 64     }
 65     int h;
 66     for(i=0;i<n;i++){
 67         if(tree[i].f<0){
 68             h=i;
 69             break;
 70         }
 71     }
 72 
 73     //cout<<h<<endl;
 74     //prefind(h);
 75 
 76     int p;
 77     queue<int> q,qq;
 78     q.push(h);
 79     while(!q.empty()){
 80         p=q.front();
 81         q.pop();
 82 
 83         //cout<<p<<endl;
 84 
 85         if(tree[p].l>=0||tree[p].r>=0){
 86             if(tree[p].l>=0)
 87                q.push(tree[p].l);
 88             if(tree[p].r>=0)
 89                q.push(tree[p].r);
 90         }
 91         else{
 92             qq.push(p);
 93         }
 94     }
 95 
 96     //cout<<h<<endl;
 97 
 98     if(!qq.empty()){
 99         printf("%d",qq.front());
100         qq.pop();
101     }
102     while(!qq.empty()){
103         printf(" %d",qq.front());
104         qq.pop();
105     }
106     printf("\n");
107     return 0;
108 }

 

posted @ 2015-08-14 08:13  Deribs4  阅读(207)  评论(0)  编辑  收藏  举报