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

生词

英文	解释
indices	index的复数

题目大意：

反转一棵二叉树，给出原二叉树的每个结点的左右孩子，输出它的层序和中序遍历～

分析：

1. 反转二叉树就是存储的时候所有左右结点都交换。
2. 二叉树使用{id, l, r, index, level}存储每个结点的id, 左右结点,下标值，和当前层数～
3. 根结点是所有左右结点中没有出现的那个结点～
4. 已知根结点，用递归的方法可以把中序遍历的结果push_back到数组v1里面，直接输出就是中序，排序输出就是层序（排序方式，层数小的排前面，相同层数时，index大的排前面）

原文链接：https://blog.csdn.net/liuchuo/article/details/52175736

题解

#include <bits/stdc++.h>

using namespace std;
const int maxn=110;
struct Node
{
    int lchild,rchild;
}node[maxn];
bool notRoot[maxn];
int n,num=0;

int strToNum(char c){
    if(c=='-') return -1;
    else{
        notRoot[c-'0']=true;
        return c-'0';
    }
}
int findRoot(){
    for(int i=0;i<n;i++){
        if(notRoot[i]==false)
            return i;
    }
}
void print(int id){
    cout<<id;
    num++;
    if(num<n) cout<<" ";
    else cout<<endl;
}
void postOrder(int root){
    if(root==-1) return;
    postOrder(node[root].lchild);
    postOrder(node[root].rchild);
    swap(node[root].lchild,node[root].rchild);
}
void BFS(int root){
    queue<int> q;
    q.push(root);
    while(!q.empty()){
        int now=q.front();
        q.pop();
        print(now);
        if(node[now].lchild!=-1) q.push(node[now].lchild);
        if(node[now].rchild!=-1) q.push(node[now].rchild);
    }
}
void inOrder(int root){
    if(root==-1) return;
    inOrder(node[root].lchild);
    print(root);
    inOrder(node[root].rchild);
}

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    char lchild,rchild;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        //getchar();
        scanf("%*c%c %c",&lchild,&rchild);
        node[i].lchild=strToNum(lchild);
        node[i].rchild=strToNum(rchild);
    }
    int root=findRoot();
    postOrder(root);
    BFS(root);
    num=0;
    inOrder(root);
    return 0;
}