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


代码：
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iomanip>

using namespace std;
struct tree
{
    int l,r;
}s[10];
int n,v[10],head,flag = 0;
char a,b;
void invert(int head)
{
    if(head == -1)return;
    invert(s[head].l);
    invert(s[head].r);
    swap(s[head].l,s[head].r);
}
void in_order(int head)
{
    if(head == -1)return;
    in_order(s[head].l);
    if(flag)printf(" %d",head);
    else printf("%d",head);
    flag ++;
    in_order(s[head].r);
}
void level_order(int head)
{
    int q[10],he = 0,ta = 0;
    if(head == -1)return ;
    q[ta ++] = head;
    while(he < ta)
    {
        int i = q[he];
        if(s[i].l != -1)q[ta ++] = s[i].l;
        if(s[i].r != -1)q[ta ++] = s[i].r;
        if(he)cout<<' '<<q[he ++];
        else cout<<q[he ++];
    }
    cout<<endl;
}
int main()
{
    cin>>n;
    for(int i = 0;i < n;i ++)
    {
        cin>>a>>b;
        if(a != '-')s[i].l = a - '0',v[s[i].l] = 1;
        else s[i].l = -1;
        if(b != '-')s[i].r = b - '0',v[s[i].r] = 1;
        else s[i].r = -1;
    }
    for(int i = 0;i < n;i ++)
    if(!v[i])
    {
        head = i;
        break;
    }
    invert(head);
    level_order(head);
    in_order(head);
}