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 (≤) 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
思路：
先找到根节点，数字r未出现，则r为根节点，因为根节点不是任何节点的子节点
然后静态构造树
再调用平常的层序遍历和中序遍历
所谓的二叉树反转，就是原来先读左子树，再读右子树
现在改为先读右子树，再读左子树
1 #include <iostream> 2 #include <vector> 3 #include <queue> 4 using namespace std; 5 struct Node 6 { 7 int l, r; 8 }; 9 int N, root[15] = { 0 }; 10 Node tree[15]; 11 vector<int>lev, in; 12 void levelOrde(int t) 13 { 14 if (t == -1) 15 return; 16 queue<int>q; 17 q.push(t); 18 while (!q.empty()) 19 { 20 t = q.front(); 21 q.pop(); 22 lev.push_back(t); 23 if (tree[t].r != -1)//先进右 24 q.push(tree[t].r); 25 if (tree[t].l != -1) 26 q.push(tree[t].l); 27 } 28 } 29 void inOrder(int t) 30 { 31 if (t == -1) 32 return; 33 inOrder(tree[t].r); 34 in.push_back(t); 35 inOrder(tree[t].l); 36 } 37 int main() 38 { 39 cin >> N; 40 char l, r; 41 for (int i = 0; i < N; ++i) 42 { 43 cin >> l >> r; 44 if (l != '-') 45 { 46 tree[i].l = l - '0'; 47 root[l - '0'] = -1;//去除为根的可能性 48 } 49 else 50 tree[i].l = -1; 51 if (r != '-') 52 { 53 tree[i].r = r - '0'; 54 root[r - '0'] = -1;//去除为根的可能性 55 } 56 else 57 tree[i].r = -1; 58 } 59 for (int i = 0; i < N; ++i) 60 { 61 if (root[i] == 0) 62 { 63 r = i; 64 break;//找到了根节点 65 } 66 } 67 levelOrde(r); 68 inOrder(r); 69 for (int i = 0; i < N; ++i) 70 cout << lev[i] << (i == N - 1 ? "" : " "); 71 cout << endl; 72 for (int i = 0; i < N; ++i) 73 cout << in[i] << (i == N - 1 ? "" : " "); 74 return 0; 75 }