1102 Invert a Binary Tree (25 分)

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

作者
CHEN, Yue
单位
浙江大学
代码长度限制
16 KB
时间限制
400 ms
内存限制
64 MB

//1102 Invert a Binary Tree (25 分)#include<bits/stdc++.h>
using namespace std;
int n;
const int    MAX_N=22;
vector<int>tree[MAX_N];
bool havaFa[MAX_N];
void level(int root){vector<int>levelArr;queue<int>q;q.push(root);while(!q.empty()){int fr= q.front();q.pop();levelArr.emplace_back(fr);if(tree[fr].size()){for(int i=tree[fr].size()-1;i>=0;i--){int val=tree[fr][i];if(val!=-1){q.push(val);}}
//            for(auto val: tree[fr]){//                q.push(val);
//            }}}int fir=1;for(auto val:levelArr){if(fir)fir=0;else printf(" ");printf("%d",val);}putchar('\n');
}
int inFir=1;
void in(int node){if(node==-1)return;
//     if(tree[node].size()>=2)in(tree[node][1]);if(inFir)inFir=0;else printf(" ");printf("%d",node);//      if(tree[node].size()>=1)in(tree[node][0]);
}
int main(){scanf("%d", &n);getchar();
//    printf("n: %d\n",n);for(int i=0;i<n;i++){char u,v;scanf("%c %c",&u,&v);getchar();if(u!='-'){tree[i].emplace_back(u-'0');havaFa[u-'0']=1;}else{//             之前没有 写push -1, 样例2 错了
//             原因是 不写的话, 会把 - 1 和 1 - 当作同一种东西tree[i].emplace_back(-1);}if(v!='-'){tree[i].emplace_back(v-'0');havaFa[v-'0']=1;}else{tree[i].emplace_back(-1);}}int root=0;for(int nodeNum=0;nodeNum<n;nodeNum++){if(!havaFa[nodeNum]){root=nodeNum;break;}}level(root);in(root);}