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
题目大意：给出一颗二叉树， 求他的镜像二叉树的层序遍历和中序遍历； 方法和一般的遍历一样， 只是先遍历右子树再遍历左子树1 #include<iostream> 2 #include<vector> 3 #include<queue> 4 using namespace std; 5 vector<vector<int> > v(10); 6 vector<int> in, exist(10, 1), level; 7 8 void inOrder(int root){ 9 if(root==-1) return;//先遍历右子树再遍历左子树 10 if(v[root][1]!=-1) inOrder(v[root][1]); 11 in.push_back(root); 12 if(v[root][0]!=-1) inOrder(v[root][0]); 13 } 14 15 int main(){ 16 int n, i, root, left, right; 17 char l, r; 18 scanf("%d", &n); 19 for(i=0; i<n; i++){ 20 cin>>l>>r; 21 left = l=='-' ? -1 : l-'0'; 22 right = r=='-' ? -1 : r-'0'; 23 if(left>=0) exist[left]=0; 24 if(right>=0) exist[right]=0; 25 v[i].push_back(left); v[i].push_back(right); 26 } 27 for(i=0; i<n; i++) if(exist[i]) root=i; 28 queue<int> q; 29 q.push(root); 30 while(q.size()){ 31 int temp=q.front(); 32 q.pop(); 33 level.push_back(temp); 34 if(v[temp][1]!=-1) q.push(v[temp][1]); 35 if(v[temp][0]!=-1) q.push(v[temp][0]); 36 } 37 inOrder(root); 38 printf("%d", level[0]); 39 for(i=1; i<n; i++) printf(" %d", level[i]); 40 printf(" %d", in[0]); 41 for(i=1; i<n; i++) printf(" %d", in[i]); 42 return 0; 43 }