1110 Complete Binary Tree

1110 Complete Binary Tree （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

解题思路：

本题考察点是判断完全二叉树。题主在调试代码时，有三个三测试点因为相同原因不能通过。节点的id可以是两位数。
补充函数：int stoi (const string& str, size_t idx = 0, int base = 10);*
Convert string to integer

判断完全二叉树（Complete Binary Tree）

若节点的左子树为空，则将节点的left设为-1，（有别于节点的编号），右子树同理，然后采用dfs算法（层序遍历），当遇到左（右子树）为空时，判断当前访问的节点数是否为n，若不为n则此树不是完全二叉树

判断满二叉树（Full Binary Tree）

与判断完全二叉树同理，判断总结点数是否是n(=2^level-1),其中level是树的深度。

#include <iostream>
#include <queue>
using namespace std;
const int maxn=200;
int left1[maxn],right1[maxn];
bool exist[maxn];
int main()
{
    int n;
    cin>>n;
    for(int i=0; i<n; i++) {
        string x1,x2;
        cin>>x1>>x2;
        if(x1=="-") left1[i]=-1;
        else {
            left1[i]=stoi(x1);
            exist[left1[i]]=true;
        }
        if(x2=="-") right1[i]=-1;
        else {
            right1[i]=stoi(x2);
            exist[right1[i]]=true;
        }
    }
    int root=0;
    while(root<n&&exist[root]) root++;
    int cnt=0;
    queue<int> q;
    q.push(root);
    bool ans=true;
    int lastnode=0;
    while(!q.empty()){
        int t=q.front();
        q.pop();
        if(t!=-1) {
            cnt++;
            lastnode=t;
        }
        else{
            if(cnt!=n){
                cout<<"NO " <<root<<endl;
            }else 
                cout<<"YES "<<lastnode<<endl;
            return 0;
        }
        
        q.push(left1[t]);
        q.push(right1[t]);
    }
    return 0;
}