1102 Invert a Binary Tree——PAT甲级

The following is from Max Howell @twitter:

Google: 90% of our engineers use the software you wrote (Homebrew), but you can't invert a binary tree on a whiteboard so fuck off.

Now it's your turn to prove that YOU CAN invert a binary tree!

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤10) 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 from 0 to N−1, 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 test case, print in the first line the level-order, and then in the second line the in-order traversal sequences of the inverted tree. There must be exactly one space between any adjacent numbers, and no extra space at the end of the line.

Sample Input:

8
1 -
- -
0 -
2 7
- -
- -
5 -
4 6

Sample Output:

3 7 2 6 4 0 5 1
6 5 7 4 3 2 0 1

 solution:
 

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl '\n'
const int N=20;
int n,head=0;
int l[N],r[N];
int hasfather[N];
vector<int>level,in;
queue<int>q;
void inorder(int n)//中序遍历
{
	if(n==-1)return;
	inorder(l[n]);
	in.push_back(n);
	inorder(r[n]);
}
void flip(int i)//反转
{
	if(i==-1)return;
	flip(l[i]);
	flip(r[i]);
    swap(l[i],r[i]);
}
int main()
{
	cin>>n;
    memset(l, -1, sizeof l);
    memset(r, -1, sizeof r);
	for(int i=0;i<n;i++)
	{
		char x,y;
		cin>>x>>y;
        //注意不要写成批注那两行，会段错误
		//l[i]=x=='-'?-1:(x-'0',hasfather[x-'0']++);
		//r[i]=y=='-'?-1:(y-'0',hasfather[y-'0']++);
        if(x!='-')l[i]=x-'0',hasfather[l[i]]=1;
        if(y!='-')r[i]=y-'0',hasfather[r[i]]=1;
	}
	while(hasfather[head])head++;//找到根节点，或者可以开一个vis数组
	flip(head);
    inorder(head);
    q.push(head);//levelorder(head);
    while(!q.empty())//运用队列进行层序遍历
    {
    	level.push_back(q.front());
    	if(l[q.front()]!=-1)q.push(l[q.front()]);
    	if(r[q.front()]!=-1)q.push(r[q.front()]);
    	q.pop();
	}
	for(int i=0;i<level.size();i++)
	{
		if(i)cout<<' ';
		cout<<level[i];
	}
	cout<<endl;
	for(int i=0;i<in.size();i++)
	{
		if(i)cout<<' ';
		cout<<in[i];
	}
	
}