PAT 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

树的层序遍历和中序遍历； 方法和普通的遍历一样，只不过这里要求翻转一棵树，只需要遍历的时候先遍历右子树，再遍历左子树就行了；
题目中没给出根节点，在输入中记录每一个出现的数字， 0~N-1中没有出现的数字就是根节点；

#include<iostream>
#include<vector>
#include<queue>
using namespace std;
vector<vector<int>> v(10);
vector<int> inorder, exist(10, 1), level;

void dfs(int root){
  if(root==-1) return;
  if(v[root][1]!=-1) dfs(v[root][1]);
  inorder.push_back(root);
  if(v[root][0]!=-1) dfs(v[root][0]);
}

int main(){
  int n, i, root, left, right;
  char l, r;
  cin>>n;
  for(i=0; i<n; i++){
    cin>>l>>r;
    left = l=='-' ? -1 : l-'0';
    right = r=='-' ? -1 : r-'0';
    if(left>=0) exist[left]=0;
    if(right>=0) exist[right]=0;
    v[i].push_back(left); v[i].push_back(right);
  }
  for(i=0; i<n; i++) if(exist[i]) root=i;
  queue<int> q;
  q.push(root);
  while(q.size()){
    int temp=q.front();
    q.pop();
    level.push_back(temp);
    if(v[temp][1]!=-1) q.push(v[temp][1]);
    if(v[temp][0]!=-1) q.push(v[temp][0]);
  }
  dfs(root);
  cout<<level[0];
  for(i=1; i<n; i++) cout<<" "<<level[i];
  cout<<endl;
  cout<<inorder[0];
  for(i=1; i<n; i++) cout<<" "<<inorder[i];
  cout<<endl;
  return 0;
}