PAT-甲级-1102 Invert a Binary Tree (25 分)

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

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;
#define rep(i,j,k)    for(int i=j;i<k;i++)
struct node{
    int id,l,r,index,level;
};
vector<node> v,in,lev;
bool cmp(node a,node b){
    if(a.level!=b.level)
        return a.level<b.level;
    return a.index > b.index;
}
void dfs(int u,int index,int level){
    if(v[u].r!=-1)
        dfs(v[u].r,2*index+2,level+1);
    node no = {u,0,0,index,level};
    in.push_back(no);
    if(v[u].l!=-1)
        dfs(v[u].l,2*index+1,level+1);
}
int main(){
    std::ios::sync_with_stdio(false);
    int n,a,b,root=0,cnt[100]={0};
    string l,r;
    cin>>n;
    v.resize(n);
    rep(i,0,n){
        cin>>l>>r;
        a = l== "-" ? -1 : stoi(l);
        b = r== "-" ? -1 : stoi(r);
        node no = {i,a,b,0,0};
        v[i] = no;
        if(a!=-1)    cnt[a]++;
        if(b!=-1)    cnt[b]++;
    }
    while(cnt[root]==1) root++;
    dfs(root,0,0);
    lev = in;
    sort(lev.begin(),lev.end(),cmp);
    rep(i,0,lev.size()){
        if(i!=0)    cout<<" ";
        cout<<lev[i].id;
    }
    cout<<endl;
    rep(i,0,in.size()){
        if(i!=0)    cout<<" ";
        cout<<in[i].id;
    }
    return 0;
}