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

解释：

给你一个二叉树

翻转这个二叉树，每个结点的左右子树交换

然后第一行输出翻转后的层序，第二行输出中序

#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
const int maxn=110;
struct node
{
    int lchild,rchild;
}Node[maxn];
bool notRoot[maxn]={false};//标记根节点数组
int n, num=0;//n为结点个数，num为已输出结点个数，控制空格输出
//输出
void print(int id)
{
    printf("%d",id);
    num++;
    if(num<n) 
        printf(" ");
    else 
        printf("\n");
}
// 中序
void inOrder(int root)
{
    if(root==-1) return;
    inOrder(Node[root].lchild);
    print(root);
    inOrder(Node[root].rchild);
}
//后序遍历，交换左右子树
void postOrder(int root)
{
    if(root==-1) 
        return;
    postOrder(Node[root].lchild);
    postOrder(Node[root].rchild);
    //交换左右孩子结点
    swap(Node[root].lchild,Node[root].rchild);
}
void BFS(int root)
{
    //注意队列存的是地址
    queue<int> q;
    //将根节点地址入队
    q.push(root);
    while(!q.empty())
    {
        int now=q.front();
        q.pop();
        print(now);
        if(Node[now].lchild!=-1)
            q.push(Node[now].lchild);
        if(Node[now].rchild!=-1)
            q.push(Node[now].rchild);
    }
}
//字符转成int，，同时标记孩子结点，为了找根结点
int strTonum(char c)
{
    if(c=='-') 
        return -1;
    else
    {
        notRoot[c-'0']=true;//标记c不是根结点
        return c-'0';
    }
}

int findRoot()
{
    for(int i=0;i<n;i++)
    {
        if(notRoot[i]==false)
            return i;//是根节点，返回i
    }
}

int main()
{
    char lchild,rchild;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%*c%c %c",&lchild,&rchild);//%*c是为了跳过回车
        Node[i].lchild=strTonum(lchild);
        Node[i].rchild=strTonum(rchild);
    }
    int root=findRoot();//获取根节点编号
    postOrder(root);//后序遍历反转二叉树
    BFS(root);//层次遍历
    num=0;
    inOrder(root);
    return 0;
}