1102. Invert a Binary Tree (25)

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

 #include<stdio.h>
 #include<string>
 #include<iostream>
 #include<string.h>
 #include<sstream>
 #include<vector>
 #include<map>
 #include<stdlib.h>
 #include<queue>
 using namespace std;

 struct node
 {
     node():l(-),r(-){}
     int l,r,id;
 };

 node Tree[];
 bool notroot[];
 bool fir = ;
 void inoder(int root)
 {
     if(Tree[root].l != -)
     {
         inoder(Tree[root].l);
     }
     if(fir)
     {
         fir = ;
         printf("%d",root);
     }
     else printf(" %d",root);
     if(Tree[root].r != -)
     {
         inoder(Tree[root].r);
     }
 }
 int main()
 {
     int n,tem;
     char l[],r[];
     scanf("%d",&n);
     for(int i = ;i <n;++i)
     {
         Tree[i].id = i;
     }
     for(int i = ;i <n;++i)
     {
         scanf("%s%s",r,l);
         if(l[] != '-')
         {
             tem = atoi(l);
             Tree[i].l = tem;
             notroot[tem] = ;
         }
         if(r[] != '-')
         {
             tem = atoi(r);
             Tree[i].r = atoi(r);
             notroot[tem] = ;
         }
     }
     int root;
     for(int i = ;i <n;++i)
     {
         if(!notroot[i])
         {
             root = i;
             break;
         }
     }
     queue<node> qq;
     qq.push(Tree[root]);
     bool fir2 = ;
     while(!qq.empty())
     {
         node ntem = qq.front();
         qq.pop();
         if(fir2)
         {
             fir2 = ;
             printf("%d",ntem.id);
         }
         else printf(" %d",ntem.id);
         if(ntem.l != -)
         {
             qq.push(Tree[ntem.l]);
         }
         if(ntem.r != -)
         {
             qq.push(Tree[ntem.r]);
         }
     }
     printf("\n");
     inoder(root);
     printf("\n");
     return ;
 }