1102. Invert a Binary Tree (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

81 -- -0 -2 7- -- -5 -4 6

Sample Output:

3 7 2 6 4 0 5 16 5 7 4 3 2 0 1

Invert  一开始没明白....

注意：用getchar()消回车

#include<stdio.h>#include<ctype.h>#include<queue>using namespace std;int n,flag=0;struct node{int left;int right;}stu[15];void in(int root){if(root!=-1){in(stu[root].left);if(flag==0){printf("%d",root);flag=1;}else{printf(" %d",root);}in(stu[root].right);}}int main(){int i;char c1,c2;scanf("%d",&n);getchar();//!!int mark[n];for(i=0;i<n;i++){mark[i]=0;}for(i=0;i<n;i++){stu[i].left=stu[i].right=-1;scanf("%c %c",&c1,&c2);if(isdigit(c2)){stu[i].left=c2-'0';mark[c2-'0']=1;}if(isdigit(c1)){stu[i].right=c1-'0';mark[c1-'0']=1;}getchar();//!!}int root;for(i=0;i<n;i++){if(mark[i]==0){root=i;break;}}queue<int>q;q.push(root);while(!q.empty()){int head=q.front();q.pop();if(flag==0){printf("%d",head);flag=1;}else{printf(" %d",head);}if(stu[head].left!=-1){q.push(stu[head].left);}if(stu[head].right!=-1){q.push(stu[head].right);}}printf("\n");flag=0;in(root);}