1110. Complete Binary Tree (25)完全二叉树

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<iostream>
#include<queue>
#include<cstring>
#include<cstdio>
using namespace std;
struct node{
int num,left,right;};
struct node N[100];
int n;
int isroot[25];
int vis[25];
int total=1;
int str_int(char a[])
{
if (a[0]=='-')
return -1;
else
{
int sum=0;
for (int i=0;i<strlen(a);i++)
sum=sum*10+a[i]-'0';
return sum;
}
}
int check()
{
for (int i=0;i<n;i++)
{
if (vis[i]==0)
return 0;
}
return 1;
}
void bfs(int root)
{
queue<int> que;
que.push(root);
int endx=root;
while(!que.empty())
{
int num=que.front();
que.pop();
if (N[num].left==-1)
{
if (total==n)
{
printf("YES %d",endx);
return ;
}
else
{
printf("NO %d",root);
return ;
}
}
else
{
que.push(N[num].left);
endx=N[num].left;
total++;
}
if (N[num].right==-1)
{
if (total==n)
{
printf("YES %d",endx);
return ;
}
else
{
printf("NO %d",root);
return ;
}
}
else
{
que.push(N[num].right);
endx=N[num].right;
total++;
}
}
}
int main()
{
int root;
char a[3],b[3];
cin>>n;
for (int i=0;i<n;i++)
{
cin>>a>>b;
int aa=str_int(a);
int bb=str_int(b);
isroot[aa]=isroot[bb]=1;
N[i].num=i;
N[i].left=aa;
N[i].right=bb;
}
for (int i=0;i<n;i++)
{
if(isroot[i]==0)
{
root=i;
break;
}
}
bfs(root);
return 0;
}