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

 1 #include<stdio.h>
 2 #include<string>
 3 #include<iostream>
 4 #include<string.h>
 5 #include<sstream>
 6 #include<vector>
 7 #include<map>
 8 #include<stdlib.h>
 9 #include<queue>
10 using namespace std;
11
12 struct node
13 {
14     node():l(-1),r(-1){}
15     int l,r,id;
16 };
17
18 node Tree[15];
19 bool notroot[15];
20 bool fir = 1;
21 void inoder(int root)
22 {
23     if(Tree[root].l != -1)
24     {
25         inoder(Tree[root].l);
26     }
27     if(fir)
28     {
29         fir = 0;
30         printf("%d",root);
31     }
32     else printf(" %d",root);
33     if(Tree[root].r != -1)
34     {
35         inoder(Tree[root].r);
36     }
37 }
38 int main()
39 {
40     int n,tem;
41     char l[5],r[5];
42     scanf("%d",&n);
43     for(int i = 0;i <n;++i)
44     {
45         Tree[i].id = i;
46     }
47     for(int i = 0;i <n;++i)
48     {
49         scanf("%s%s",r,l);
50         if(l[0] != ‘-‘)
51         {
52             tem = atoi(l);
53             Tree[i].l = tem;
54             notroot[tem] = 1;
55         }
56         if(r[0] != ‘-‘)
57         {
58             tem = atoi(r);
59             Tree[i].r = atoi(r);
60             notroot[tem] = 1;
61         }
62     }
63     int root;
64     for(int i = 0;i <n;++i)
65     {
66         if(!notroot[i])
67         {
68             root = i;
69             break;
70         }
71     }
72     queue<node> qq;
73     qq.push(Tree[root]);
74     bool fir2 = 1;
75     while(!qq.empty())
76     {
77         node ntem = qq.front();
78         qq.pop();
79         if(fir2)
80         {
81             fir2 = 0;
82             printf("%d",ntem.id);
83         }
84         else printf(" %d",ntem.id);
85         if(ntem.l != -1)
86         {
87             qq.push(Tree[ntem.l]);
88         }
89         if(ntem.r != -1)
90         {
91             qq.push(Tree[ntem.r]);
92         }
93     }
94     printf("\n");
95     inoder(root);
96     printf("\n");
97     return 0;
98 }