PAT A1110 Complete Binary Tree （25 分）——完全二叉树，字符串转数字

Given a tree, you are supposed to tell if it is a complete binary tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤20) 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 case, print in one line YES and the index of the last node if the tree is a complete binary tree, or NO and the index of the root if not. There must be exactly one space separating the word and the number.

Sample Input 1:

9
7 8
- -
- -
- -
0 1
2 3
4 5
- -
- -

Sample Output 1:

YES 8

Sample Input 2:

8
- -
4 5
0 6
- -
2 3
- 7
- -
- -

Sample Output 2:

NO 1

 #include <stdio.h>
 #include <algorithm>
 #include <set>
 #include <string.h>
 #include <vector>
 #include <math.h>
 #include <queue>
 #include <iostream>
 #include <string>
 using namespace std;
 const int maxn = ;
 int n,k;
 struct node{
     int l=-,r=-;
 }tree[maxn];
 int vis[maxn]={};
 int main(){
     scanf("%d",&n);
     getchar();
     for(int i=;i<n;i++){
         string c1,c2;
         cin>>c1>>c2;
         getchar();
         int n1=-,n2=-;
         if(c1!="-"){
             n1=stoi(c1);
             vis[n1]=;
         }
         if(c2!="-"){
             n2=stoi(c2);
             vis[n2]=;
         }
         tree[i].l=n1;
         tree[i].r=n2;
     }
     int root=-;
     for(int i=;i<n;i++){
         if(vis[i]==){
             root=i;
             break;
         }
     }
     int flag=,flag2=,now;
     queue<int> q;
     q.push(root);
     while(!q.empty()){
         now=q.front();
         q.pop();
         //printf("%d ",now);
         if(tree[now].l!=-){
             if(flag==)q.push(tree[now].l);
             else {
                 flag2=;
                 break;
             }
         }
         else{
             flag=;
         }
         if(tree[now].r!=-){
             if(flag==)q.push(tree[now].r);
             else {
                 flag2=;
                 break;
             }
         }
         else{
             flag=;
         }
         /*if(now!=-1){
             flag++;
             flag2=now;
         }
         else{
             if(flag==n){
                 printf("YES %d",flag2);
             }
             else{
                 printf("NO %d",root);
             }
             return 0;
         }
         q.push(tree[now].l);
         q.push(tree[now].r);
         */
     }
     if(flag2==)printf("NO %d",root);
     else printf("YES %d",now);
 }

注意点：完全二叉树的判断就是看已经有节点没孩子了，后面节点却还有孩子，这树就不是完全二叉树。有三个点一直错误，一开始段错误，后来在结构体初始化了-1后答案错误，总找不到原因。后来才发现原来是读输入的时候错了，两位数就读不到了，不能用%c，要用string