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

3 7 2 6 4 0 5 1
6 5 7 4 3 2 0 1

## 解答

1. 没有在输入数据中出现的编号就是根结点的编号，如样例的根结点就为3。
2. 任意选择一个编号，递归搜索，当它不是任意结点的孩子时，它就是根结点。
第一种方法比较简单，也容易实现，不容易出错。
找到根结点之后，就可以反转二叉树了，核心算法就是先交换两个子孩子，然后继续反转两个孩子结点，因为它们也是二叉树。注意孩子结点的编号不是2i也不是2i+1，因为结点序号是无序的，要根据结构体成员来获取孩子编号。
反转二叉树后，就可以使用层次遍历和中序遍历代码输出结果了。
#include<iostream>
#include<queue>

using namespace std;

struct node {
int lchild, rchild;
}nodes[15];
int count1 = 0;
int findRoot(int j, int n) {
for (int i = 0; i < n; i++) {
if (i == j) continue;
if (nodes[i].lchild == j || nodes[i].rchild == j)
return findRoot(i, n);
}
return j;
}

void levelOrder(int root)
{
queue<int> q;
q.push(root);
while (!q.empty()) {
int t = q.front();
q.pop();
if (nodes[t].lchild != -1)
q.push(nodes[t].lchild);
if (nodes[t].rchild != -1)
q.push(nodes[t].rchild);
printf("%d", t);
if (q.empty())
printf("\n");
else
printf(" ");
}
}

void invertTree(int root) {
if (root!= -1) {
swap(nodes[root].lchild, nodes[root].rchild);
invertTree(nodes[root].lchild);
invertTree(nodes[root].rchild);
}
}

void inOrder(int root, int n) {
if (root!= -1) {
inOrder(nodes[root].lchild, n);
printf("%d", root);
count1++;
if (n > count1) printf(" ");
inOrder(nodes[root].rchild, n);
}
}

int main()
{
int n;
char a[4], b[4];
cin >> n;
for (int i = 0; i < n; i++){
scanf("%s %s", a, b);
if (a[0] == '-') nodes[i].lchild = -1;
else nodes[i].lchild = atoi(a);
if (b[0] == '-') nodes[i].rchild = -1;
else nodes[i].rchild = atoi(b);
}
int root = findRoot(0, n);
invertTree(root);
levelOrder(root);
inOrder(root, n);
return 0;
}