1102 Invert a Binary Tree (25 分)

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

代码： 

#include
#include
using namespace std;
struct node
{
    int lchild,rchild;
}Node[15];
int le=0,in=0,N;
void levelorder(int root)
{
    queue Q;
    if(root==-1)
        return;
    Q.push(root);
    while(!Q.empty())
    {
        int temp=Q.front();
        Q.pop();
        printf("%d",temp);
        le++;
        if(le!=N)
            printf(" ");
        if(Node[temp].lchild!=-1)
            Q.push(Node[temp].lchild);
        if(Node[temp].rchild!=-1)
            Q.push(Node[temp].rchild);
    }
    printf("\n");
}
void inorder(int root)
{
    if(root==-1)
        return;
    inorder(Node[root].lchild);
    printf("%d",root);
    in++;
    if(in!=N)
        printf(" ");
    inorder(Node[root].rchild);
}
int main()
{
    bool  appear[10]={false};
    scanf("%d",&N);
    char c1,c2;
    for(int i=0;i='0'&&c1<='9')
        {
            appear[c1-'0']=true;
            Node[i].rchild=c1-'0';
        }
        else
            Node[i].rchild=-1;
        if(c2>='0'&&c2<='9')
        {
            appear[c2-'0']=true;
            Node[i].lchild=c2-'0';
        }
        else
            Node[i].lchild=-1;
    }
    int root;
    for(int i=0;i