原题干：
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
生词：
Invert：反转 hence：从此，因此
corresponds：相当，相符合
indices：指数
大意：
将二叉树所有节点依次编号，0,1,2,3.....N
第一行给出节点个数，在下面的几行中，依次给出每个节点的左子树和右子树的编号
顺序按照编号顺序。
题目中样例输入意思就是给出8个节点，编号0,1,2,3,4,5,6,7
第0个节点的左子树为第一个节点，右子树为NULL
第1个节点的左子树和右子树均为NULL
第二个节点左子树为0，右子树为NULL。。。
然后把二叉树所有节点的左右子树调换顺序，输出二叉树的层序排列和中序排列序列，末尾不能有多余的空格。
这道题可以使用静态二叉树实现，将数组当做内存，下标当做二叉树的编号，使用struct存储二叉树节点。使用-1表示NULL
代码如下：
#include <iostream>
#include <stack>
#include <string>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;
struct node{
int lchild, rchild;
};
node RAM[12];
bool not_root[12] = {false};
int cnt, num;
/**
* 字符转数字，如果是‘-’，返回-1
*/
int StrToNum(char c){
if(c != '-') not_root[c-'0'] = true;
return c == '-' ? -1 : (c - '0');
}
/**
* 寻找根节点
*/
int find_root(){
for(int i = 0; i < cnt; i++)
if(!not_root[i]) return i;
//return -1;
}
/**
* 二叉树的层序遍历
*/
void BFS(int root){
queue<int> Q;
Q.push(root);
while(!Q.empty()){
int now = Q.front();
Q.pop();
printf("%d", now);
if(--num) printf(" ");
if(RAM[now].lchild != -1) Q.push(RAM[now].lchild);
if(RAM[now].rchild != -1) Q.push(RAM[now].rchild);
}
}
/**
* 二叉树的中序排列
*/
void inOrder(int root){
if(root == -1) return;
inOrder(RAM[root].lchild);
printf("%d", root);
if(--num) printf(" ");
inOrder(RAM[root].rchild);
}
int main(){
char lchild, rchild;
cin >> cnt;
for(int i = 0; i < cnt; i++){
cin >> lchild >> rchild;
RAM[i].rchild = StrToNum(lchild),
RAM[i].lchild = StrToNum(rchild);
}
int root = find_root();
num = cnt;
BFS(root);
cout << endl;
num = cnt;
inOrder(root);
return 0;
}