1102. Invert a Binary Tree (25)

The following is from Max Howell @twitter:

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

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

Input Specification:

Each input file contains the one test case. For each case, the first line gives a positive integer N (<=10) which was the total number of nodes in the tree--and H Ence The nodes is 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 Node. If the child does not exist, a "-" 'll be put on the position. Any pair of children is separated by a space.

Output Specification:

For each test case, print in the first line of the Level-order, and then on the second line the In-order traversal sequences of the inverted tree. There must is exactly one space between any adjacent numbers, and no extra space at the end of the line.

Sample Input:

Bayi---0-2 7----5-4 6

Sample Output:

3 7 2 6 4 0 5 16 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>Ten using namespacestd; One  A structnode - { -Node (): L (-1), R (-1){} the     intL,r,id; - }; -  -Node tree[ the]; + BOOLnotroot[ the]; - BOOLFIR =1; + voidInoder (introot) A { at     if(TREE[ROOT].L! =-1) -     { - Inoder (TREE[ROOT].L); -     } -     if(FIR) -     { inFIR =0; -printf"%d", root); to     } +     Elseprintf"%d", root); -     if(TREE[ROOT].R! =-1) the     { * Inoder (TREE[ROOT].R); $     }Panax Notoginseng } - intMain () the { +     intN,tem; A     Charl[5],r[5]; thescanf"%d",&n); +      for(inti =0; I <n;++i) -     { $Tree[i].id =i; $     } -      for(inti =0; I <n;++i) -     { thescanf"%s%s", r,l); -         if(l[0] !='-')Wuyi         { theTEM =atoi (l); -TREE[I].L =tem; WuNotroot[tem] =1; -         } About         if(r[0] !='-') $         { -TEM =atoi (r); -TREE[I].R =atoi (r); -Notroot[tem] =1; A         } +     } the     intRoot; -      for(inti =0; I <n;++i) $     { the         if(!Notroot[i]) the         { theRoot =i; the              Break; -         } in     } theQueue<node>QQ; the Qq.push (Tree[root]); About     BOOLFir2 =1; the      while(!qq.empty ()) the     { theNode Ntem =Qq.front (); + Qq.pop (); -         if(FIR2) the         {BayiFir2 =0; theprintf"%d", ntem.id); the         }  -         Elseprintf"%d", ntem.id); -         if(NTEM.L! =-1) the         { the Qq.push (TREE[NTEM.L]); the         } the         if(NTEM.R! =-1) -         { the Qq.push (TREE[NTEM.R]); the         } the     }94printf"\ n"); the Inoder (root); theprintf"\ n"); the     return 0;98}

1102. Invert a Binary Tree (25)