1102 Invert a Binary Tree

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<iostream>
#include<vector>
#include<queue>
using namespace std;
vector<int> inorder;
vector<int>levelorder;
struct node
{
    int leftchild=-1;
    int rightchild=-1;
    int father=-1;
}Node[15];
void myinorder(int root)
{
    if(root==-1)return;
    myinorder(Node[root].leftchild);
    inorder.push_back(root);
    myinorder(Node[root].rightchild);
}
void layerorder(int root)
{
    queue<int> q;
    q.push(root);
    while(!q.empty())
    {
        int now=q.front();
        q.pop();
        levelorder.push_back(now);
        if(Node[now].leftchild!=-1) q.push(Node[now].leftchild);
        if(Node[now].rightchild!=-1) q.push(Node[now].rightchild);
    }
}
int main()
{
    int num;
    char temp;
//    scanf("%d",&num);
    cin>>num;    
    for(int i=0;i<num;i++)
    {
        for(int j=0;j<2;j++)
        {
        cin>>temp;
        //    scanf("%c",&temp);
            if(temp!='-' )
            {
                if(j==1)
                {
                    Node[i].leftchild=temp-'0';
                    Node[(temp-'0')].father=i;
                }
                else
                {
                    Node[i].rightchild=temp-'0';
                    Node[(temp-'0')].father=i;
                }
            
            }
            
        }
    
    }
    int flag=0;
    while(Node[flag].father!=-1)
    {
        flag=Node[flag].father;
    }
    myinorder(flag);
    layerorder(flag);
    int countlay=0;
    while(countlay!=num)
    {
        if(countlay==0)
        {
            printf("%d",levelorder[countlay]);
        }
        else
        {
            printf(" %d",levelorder[countlay]);
        }
        countlay++;    
    }
    printf("\n");
    int count=0;
    while(count!=num)
    {
        if(count==0)
        {
            printf("%d",inorder[count]);
        }
        else
        {
            printf(" %d",inorder[count]);
        }
        count++;    
    }
    
 }