PAT 甲级 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 (≤) 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

题意：给出 n 个结点,（0-n-1）n 行左右孩子结点，根结点为行中没有出现的结点；

判断是否为完全二叉树，是则输出 (YES) 与末尾结点，否则输出(NO)与根结点。

AC【判断完全二叉树】：1、[ 找出根结点 ] 　　 2、[ DFS遍历 ]　　 3、[ 分类输出 ]

#include<bits/stdc++.h>
using namespace std;

struct node{
    string left, right;
} Node[21];

int root= 0, maxn = -1, last;

void DFS(int root, int index)
{
    if(index > maxn)
    {
        maxn = index;
        last = root;
    }
    if(Node[root].left != "-") DFS(stoi(Node[root].left), index * 2);
    if(Node[root].right != "-") DFS(stoi(Node[root].right), index * 2 + 1);
}

int main()
{
    int n; cin >> n;
    vector<int> find_root(n, 0);
    for(int i = 0; i < n; i++)
    {
        cin >> Node[i].left >> Node[i].right;
        if(Node[i].left != "-") find_root[stoi(Node[i].left)] = 1;
        if(Node[i].right != "-") find_root[stoi(Node[i].right)] = 1;
    }
    
    while(root < n && find_root[root]) root++;
    DFS(root, 1);
    
    if(n == maxn) cout << "YES " << last;
    else cout << "NO " << root;
    return 0;
}

Tips：vector<int> v(n, 0);　　// 初始化