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：二叉树转换
#include <iostream> #include <string> #include <cstdio> #include <cmath> #include <queue> #include <vector> #include <functional> #define rep(i,j,k) for(int i=j;i<=k;++i) const int Max=10; int c[Max]; using namespace std; struct node { int left; int right; }; queue<int> q; int n; node a[Max]; void invert(int r) { if (r!=-1){ swap(a[r].left,a[r].right); invert(a[r].left); invert(a[r].right); } } void level(int r) { q.push(r); while (!q.empty()){ r=q.front(); q.pop(); if (a[r].left!=-1) q.push(a[r].left); if (a[r].right!=-1) q.push(a[r].right); cout<<r; if (!q.empty()) cout<<' '; } cout<<endl; } void inorder(int r) { if (a[r].left!=-1) inorder(a[r].left); q.push(r); if (a[r].right!=-1) inorder(a[r].right); } void in_print(int root) { inorder(root); while (!q.empty()){ cout<<q.front(); q.pop(); if (!q.empty()) cout<<' '; } cout<<endl; } int main() { // freopen("test.txt","r",stdin); cin>>n; char x,y; rep(i,0,n-1){ cin>>x>>y; if (x=='-') a[i].left=-1; else { a[i].left=x-'0'; c[x-'0']=1; } if (y=='-') a[i].right=-1; else { a[i].right=y-'0'; c[y-'0']=1; } } int root=0; rep(i,0,n-1){ if (c[i]!=1) { root=i; break; } } invert(root); level(root); in_print(root); return 0; }
