PAT A1110. Complete Binary Tree (25)

1. Complete Binary Tree (25)

时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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

感谢黄总不吝赐教。先膜为敬

   #include<cstdio>
#include<queue>
#include<algorithm>
#include<cstdlib>
using namespace std;
const int maxn=21;
bool isRoot[maxn];
struct node{
    int l,r;
}bst[maxn];
int n;
int root;
void levelorder(int r){
    queue<int>q;
    q.push(r);
    bool iscomplete=true;
    bool flag=false;
    int cnt=0;
    int last=-1;
    while(q.size()!=0){
        int x=q.front();
        q.pop();
        cnt++;
        if(cnt==n)last=x;
        if(flag==false){
            if(bst[x].l==-1&&bst[x].r!=-1){
                iscomplete=false;
            }else if(bst[x].l==-1||bst[x].r==-1){
                flag=true;
            }
        }else{
            if(bst[x].l!=-1||bst[x].r!=-1){
                iscomplete=false;
            }
        }
        if(bst[x].l!=-1)q.push(bst[x].l);
        if(bst[x].r!=-1)q.push(bst[x].r);
    }
    if(iscomplete==true)printf("YES %d\n",last);
    else printf("NO %d\n",root);
}
int main(){
    fill(isRoot,isRoot+maxn,true);
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        char a[3]="\0",b[3]="\0";
        getchar();
        scanf("%s %s",a,b);
        if(a[0]=='-')bst[i].l=-1;
        else{
            bst[i].l=atoi(a);
            isRoot[bst[i].l]=false;
        }
        if(b[0]=='-')bst[i].r=-1;
        else{
            bst[i].r=atoi(b);
            isRoot[bst[i].r]=false;
        }
    }
    for(int i=0;i<n;i++){
        if(isRoot[i]==true){
            root=i;
            break;
        }
    }
    levelorder(root);
    return 0;
}