PAT 1110 Complete Binary Tree (25)

1110 Complete Binary Tree (25)（25 分）

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

题目大意：给出一颗二叉树的结构，判断该二叉树是否是完全二叉树；
思路：用数组保存树的结构
     在节点中未出现的值就是根节点
     层序遍历二叉树， 如果已经出现一个空节点， 如果还出现非空节点， 则该树一定不是完全二叉树；
注意点：开始看输入样例， 误以为输入都是一个字符串，导致三个测试点不能通过， 找了很久的原因， 最多有20个节点， 用string输入就行

#include<iostream>
#include<vector>
#include<queue>
using namespace std;
struct node{ int left, right;};
int main(){
  int n, root, i;
  scanf("%d", &n);
  vector<node> v(n);
  //记录出现过的节点
  vector<int> vis(n, 0);
  string a, b;
  for(i=0; i<n; i++){
    cin>>a>>b;
    if(a=="-") v[i].left = -1;
    else{
      v[i].left = stoi(a);
      vis[v[i].left] = 1;
    }
    if(b=="-") v[i].right = -1;
    else{
      v[i].right = stoi(b);
      vis[v[i].right] = 1;
    }
  }
  for(i=0; i<n; i++) //没有出现过在子节点中的点就是根节点
    if(vis[i]==0) root=i; 
  queue<int> q;
  q.push(root);
  int temp, flag=0, f=0 ;
  //flag:记录是否出现空节点
  //f：记录出现空节点后， 是否还有后继的节点
  while(q.size()){
    temp = q.front();
    q.pop();
    if(f) break;
      if(v[temp].left != -1){
        q.push(v[temp].left);
        if(flag) f=1;
      }else flag=1;
      if(v[temp].right != -1){
        q.push(v[temp].right);
        if(flag) f=1;
      }else flag=1;
  }
  if(f) printf("NO %d", root);
  else printf("YES %d", temp);
  return 0;
}