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

分析：根據輸入建立一棵樹，然後倒轉樹，即結點的左右子樹互相交換。 使用陣列儲存樹節點，然後層次遍歷和中序遍歷即可。 參考程式碼：

#include<iostream>
#include<vector>
#include<algorithm>
#include<queue>
using namespace std;
struct node {
    int left;
    int right;
};
int flag = 0;
vector<node>v;
void in_order(int vec) {
    if (v[vec].left != -1)
        in_order(v[vec].left);
    if (flag)
        cout << " ";
    cout << vec;
    flag = 1;
    if (v[vec].right != -1)
        in_order(v[vec].right);
}
int main()
{
    int n;
    char ch1, ch2;
    scanf_s("%d", &n);
    v.resize(n);
    vector<int>num(n, -1);
    getchar();
    for (int i = 0; i < n; i++) {
        ch1 = getchar();
        getchar();
        ch2 = getchar();
        getchar();
        if (ch1 == '-')
            v[i].right = -1;
        else { v[i].right = ch1 - '0'; num[ch1 - '0'] = 1; }
        if (ch2 == '-')
            v[i].left = -1;
        else { v[i].left = ch2 - '0'; num[ch2 - '0'] = 1; }
    }
    int root;
    for (int i = 0; i < n; i++)
    {
        if (num[i] == -1) {
            root = i;
            break;
        }
    }

    int flag = 0;
    queue<int>q;
    q.push(root);
    while (!q.empty())
    {
        int temp = q.front();
        if (flag)
            cout << " ";
        cout << temp;
        flag = 1;
        q.pop();
        if (v[temp].left != -1)
            q.push(v[temp].left);
        if (v[temp].right != -1)
            q.push(v[temp].right);
    }

    cout << endl;
    in_order(root);


    return 0;
}