PAT Advanced 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 (≤) 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

This problem is in turn given node left child and right child, were turned over to find the binary tree traversal sequence and preorder.

We can use the build tree traversal sequence, followed by using the sort carried out according to level from small to large, from largest to smallest conducted according to index. Can be obtained reversing the binary tree traversal sequence.

The last print

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;
struct node {
    int index, id, left = -1, right = -1, level;
}a[100];
int N;
vector<node> v;
bool cmp(node& n1, node& n2){
    return n1.level == n2.level ? n1.index > n2.index: n1.level < n2.level;
}
void dfs(int root, int index, int level){
    if(a[root].right != -1) dfs(a[root].right, index * 2 + 2, level + 1);
    v.push_back({index, root, 0, 0, level});
    if(a[root].left != -1) dfs(a[root].left, index * 2 + 1, level + 1);
}
int main () {
    cin >> N;
    string tmp_l, tmp_r;
    vector<bool> find_root(N);
    for(int i = 0; i < N; i++){
        cin >> tmp_l >> tmp_r;
        a[i].id = i;
        if(tmp_l != "-") {
            a[i].left = stoi(tmp_l);
            find_root[stoi(tmp_l)] = true;
        }
        if(tmp_r != "-") {
            a[i].right = stoi(tmp_r);
            find_root[stoi(tmp_r)] = true;
        }
    }
    int root_index;
    for(int i = 0; i < N; i++)
        if(!find_root[i]) root_index = i;
    dfs(root_index, 0, 0);
    vector<node> v2(v);
    sort(v2.begin(), v2.end(), cmp);
    for(int i = 0; i < N; i++)
        if(i != N-1) cout << v2[i].id << " ";
        else cout << v2[i].id <<endl;
    for(int i = 0; i < N; i++)
        if(i != N-1) cout << v[i].id << " ";
        else cout << v[i].id <<endl;
    system("pause");
    return 0;
}