PAT (Advanced Level) Practice 1110 Complete Binary Tree (25分) （完全二叉树的判断+分享致命婴幼儿错误）

1.题目

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

2.题目分析

最通俗的讲，完全二叉树其实就是层序遍历时，必须从左到右节点都不是NULL，直至结束。如果存在一个节点为NULL，而之后又有非NULL的节点，就不是二叉树。

于是我的判断方法就是即使节点是NULL也放入队列，再输出的时候遇到NULL判断已经输出的个数是不是总节点的个数，不是的话就不是完全二叉树。

使用20之后的一个位置存放-1，表示NULL

致命婴幼儿错误：节点编号可能为两位数………………

 3.代码

#include<iostream>
#include<queue>
#include<string>
#include<cstring>
using namespace std;
struct node
{
	int data;
	int left;
	int right;
}list[30],output,output2;
int mark[30];

int counts = 0;
bool ok = true;
int judge(int head,int n)
{
	queue<node>out;
	out.push(list[head]);
	while (!out.empty())
	{
		output = out.front();
		out.pop();
		if (output.data == -1)
		{
			if (counts < n) { ok = false; return -1; }
			else return output2.data;
		}
		counts++;
	out.push(list[output.left]);	
	out.push(list[output.right]);
	output2 = output;//记录上一个output的值，当最后遍历完后，output变为最后的NULL（-1），output2就是所求的最后一个节点的值
	}
}

int main()
{
	int n,head;
	string a, b;
	list[25].data = -1;
	scanf("%d", &n);
	for (int i = 0; i < n; i++)
	{
		list[i].data = i;
		cin >> a >> b;
		if (a != "-") { list[i].left = stoi(a); mark[list[i].left] = 1; }
		else list[i].left = 25;
		if (b != "-") { list[i].right = stoi(b); mark[list[i].right] = 1;}
		else list[i].right = 25;
	}
	for (int i = 0; i < n; i++)
		if (mark[i] == 0)head = i;
	int c = judge(head, n);
	if (ok)printf("%s %d\n", "YES", c);
	else printf("%s %d\n", "NO", list[head]);
}