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 ;
}

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