PAT甲级——A1110 Complete Binary Tree【25】

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

 #include <iostream>
 #include <vector>
 #include <queue>
 #include <string>
 using namespace std;
 struct Node
 {
     int l, r;
 }node;
 int n;
 int main()
 {
     cin >> n;
     vector<Node>tree;
     vector<bool>isRoot(n, true);
     for (int i = ; i < n; ++i)
     {
         string s1, s2;
         cin >> s1 >> s2;
         if (s1 == "-")
             node.l = -;
         else
         {
             node.l = atoi(s1.c_str());
             isRoot[node.l] = false;
         }
         if (s2 == "-")
             node.r = -;
         else
         {
             node.r = atoi(s2.c_str());
             isRoot[node.r] = false;
         }
         tree.push_back(node);
     }
     int root = -;//根
     for (int i = ; i < n && root==-; ++i)
         if (isRoot[i])
             root = i;
     queue<int>q, temp;
     q.push(root);
     while (!q.empty())//进行层序遍历
     {
         int p = q.front();
         q.pop();
         temp.push(p);//保存弹出的数据
         if (tree[p].l != -)
             q.push(tree[p].l);
         else
             break;//出现空子节点，则打破了完全二叉树的规则
         if (tree[p].r != -)
             q.push(tree[p].r);
         else
             break;//出现空子节点，则打破了完全二叉树的规则
     }
     if (temp.size() + q.size() == n)//满足完全二叉树的要求
         cout << "YES " << q.back() << endl;//最后压入的就是最后一个节点
     else
         cout << "NO " << root << endl;
     return ;
 }