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
#include<iostream> #include<vector> #include<queue> using namespace std; vector<int> inorder; vector<int>levelorder; struct node { int leftchild=-1; int rightchild=-1; int father=-1; }Node[15]; void myinorder(int root) { if(root==-1)return; myinorder(Node[root].leftchild); inorder.push_back(root); myinorder(Node[root].rightchild); } void layerorder(int root) { queue<int> q; q.push(root); while(!q.empty()) { int now=q.front(); q.pop(); levelorder.push_back(now); if(Node[now].leftchild!=-1) q.push(Node[now].leftchild); if(Node[now].rightchild!=-1) q.push(Node[now].rightchild); } } int main() { int num; char temp; // scanf("%d",&num); cin>>num; for(int i=0;i<num;i++) { for(int j=0;j<2;j++) { cin>>temp; // scanf("%c",&temp); if(temp!='-' ) { if(j==1) { Node[i].leftchild=temp-'0'; Node[(temp-'0')].father=i; } else { Node[i].rightchild=temp-'0'; Node[(temp-'0')].father=i; } } } } int flag=0; while(Node[flag].father!=-1) { flag=Node[flag].father; } myinorder(flag); layerorder(flag); int countlay=0; while(countlay!=num) { if(countlay==0) { printf("%d",levelorder[countlay]); } else { printf(" %d",levelorder[countlay]); } countlay++; } printf("\n"); int count=0; while(count!=num) { if(count==0) { printf("%d",inorder[count]); } else { printf(" %d",inorder[count]); } count++; } }
标签:Node,Binary,temp,int,countlay,Invert,1102,now,root
From: https://www.cnblogs.com/zzzlight/p/16905360.html