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

Analysis

题意大概如下：第一个数字N表示有N个节点。接下来N行表示第i行的左右子节点，如果是空就用 ‘-’ 表示。然后讲左右子节点反转，输出层级遍历的结果与中序遍历的结果。

这次的代码我选择最熟悉的写法，所以不太漂亮，但是成功AC，也让我收获了层级遍历的写法，文尾有柳神的代码。

Code(C++)

#include <iostream>
#include <vector>
#include <queue>
using namespace std;

typedef struct node *PtrToNode;
typedef PtrToNode Tree;

struct node {
    int id;
    PtrToNode Left;
    PtrToNode Right;
};

vector<int> in, level;
queue<PtrToNode> que;

void inorder(Tree T)
{
    if(T == NULL)
        return;
    
    inorder(T->Left);
    in.push_back(T->id);
    inorder(T->Right);
}

void levelorder(Tree T)
{
    if (T == NULL)
        return;
    
    PtrToNode ptr = T;
    que.push(ptr);
    while (!que.empty())
    {
        ptr = que.front();
        que.pop();
        level.push_back(ptr->id);
        if (ptr->Left != NULL)
            que.push(ptr->Left);
        if (ptr->Right != NULL)
            que.push(ptr->Right);
    }
}

int main()
{
    int n;
    cin >> n;
    
    bool flag[n];
    PtrToNode a[n];
    for(int i=0; i<n; i++)
    {
        a[i] = (PtrToNode)malloc(sizeof(struct node));
        a[i]->id = i;
        a[i]->Left = a[i]->Right = NULL;
        flag[i] = true;
    }
    
    
    for(int i=0; i<n; i++)
    {
        char l, r;
        cin >> l >> r;
        if(l != '-')
        {
            a[i]->Right = a[l-'0'];
            flag[l -'0'] = false;
        }
        if(r != '-')
        {
            a[i]->Left = a[r-'0'];
            flag[r-'0'] = false;
        }
    }
    
    Tree T;
    for(int i=0; i<n; i++)
        if(flag[i])
        {
            T = a[i];
            break;
        }
    
    levelorder(T);
    cout << level[0];
    for(int i=1; i<level.size(); i++)
        cout << " " << level[i];
    
    cout << endl;
    
    inorder(T);
    cout << in[0];
    for(int i=1; i<in.size(); i++)
        cout << " " << in[i];
}

柳神的代码，还是一如即往地选择vector来建树。

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
struct node {
    int id, l, r, index, level;
} a[100];
vector<node> v1;
void dfs(int root, int index, int level) {
    if (a[root].r != -1) dfs(a[root].r, index * 2 + 2, level + 1);
    v1.push_back({root, 0, 0, index, level});
    if (a[root].l != -1) dfs(a[root].l, index * 2 + 1, level + 1);
}
bool cmp(node a, node b) {
    if (a.level != b.level) return a.level < b.level;
    return a.index > b.index;
}
int main() {
    int n, have[100] = {0}, root = 0;
    cin >> n;
    for (int i = 0; i < n; i++) {
        a[i].id = i;
        string l, r;
        cin >> l >> r;
        if (l != "-") {
            a[i].l = stoi(l);
            have[stoi(l)] = 1;
        } else {
            a[i].l = -1;
        }
        if (r != "-") {
            a[i].r = stoi(r);
            have[stoi(r)] = 1;
        } else {
            a[i].r = -1;
        }
    }
    while (have[root] == 1) root++;
    dfs(root, 0, 0);
    vector<node> v2(v1);
    sort(v2.begin(), v2.end(), cmp);
    for (int i = 0; i < v2.size(); i++) {
        if (i != 0) cout << " ";
        cout << v2[i].id;
    }
    cout << endl;
    for (int i = 0; i < v1.size(); i++) {
        if (i != 0) cout << " ";
        cout << v1[i].id;
    }
    return 0;
}