PAT 1102 Invert a Binary Tree[比较简单]

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

 题目大意：给出一颗二叉树，节点数<=10，可以说很少了，将其左右反转，就是原来的左子树变为右子树，递归进行，并且输出反转后的层次遍历和中序遍历。

我的代码：应该是正确的，但是在层次遍历中因为使用了递归，所以不好控制最后的空格，所以全部测试点格式错误0分，也不能通过传参标记来控制吧。那么也就是说不能通过递归来进行了？ 

#include <iostream>
#include <map>
#include <cstdio>
#include <queue>
using namespace std;
struct Node{
    int father;
    int left,right;
    Node(){
        left=-1;right=-1;father=-1;
    }
}node[11];
int root;
void inorder(int r){//但是这个怎么去控制最后一个不输出空格呢？哭唧唧啊。
    if(node[r].left!=-1)
        inorder(node[r].left);
    cout<<r<<" ";
    if(node[r].right!=-1)
        inorder(node[r].right);
}
int main() {
    int n;
    cin>>n;
    char ch1,ch2;
    for(int i=0;i<n;i++){
        cin>>ch1>>ch2;
        if(ch1!='-'){
            node[i].right=ch1-'0';
            node[ch1-'0'].father=i;
        }
        if(ch2!='-'){
            node[i].left=ch2-'0';
            node[ch2-'0'].father=i;
        }
    }
    root=-1;
    for(int i=0;i<n;i++){
        if(node[i].father==-1){
            root=i;break;
        }
    }
    //层次遍历的结果
    queue<int> que;//现在完全不知道根是哪一个。
    que.push(root);
    while(!que.empty()){
        int top=que.front();
        que.pop();
        cout<<top;
        if(node[top].left!=-1)que.push(node[top].left);
        if(node[top].right!=-1)que.push(node[top].right);
        if(!que.empty())cout<<" ";
    }
    cout<<endl;
    inorder(root);

    return 0;
}

#include <iostream>
#include <map>
#include <cstdio>
#include <queue>
using namespace std;
struct Node{
    int father;
    int left,right;
    Node(){
        left=-1;right=-1;father=-1;
    }
}node[11];
int root;
vector<int> in;
void inorder(int r){//但是这个怎么去控制最后一个不输出空格呢？哭唧唧啊。
    if(node[r].left!=-1)
        inorder(node[r].left);
    in.push_back(r);
    if(node[r].right!=-1)
        inorder(node[r].right);
}
int main() {
    int n;
    cin>>n;
    char ch1,ch2;
    for(int i=0;i<n;i++){
        cin>>ch1>>ch2;
        if(ch1!='-'){
            node[i].right=ch1-'0';
            node[ch1-'0'].father=i;
        }
        if(ch2!='-'){
            node[i].left=ch2-'0';
            node[ch2-'0'].father=i;
        }
    }
    root=-1;
    for(int i=0;i<n;i++){
        if(node[i].father==-1){
            root=i;break;
        }
    }
    //层次遍历的结果
    queue<int> que;//现在完全不知道根是哪一个。
    que.push(root);
    while(!que.empty()){
        int top=que.front();
        que.pop();
        cout<<top;
        if(node[top].left!=-1)que.push(node[top].left);
        if(node[top].right!=-1)que.push(node[top].right);
        if(!que.empty())cout<<" ";
    }
    cout<<endl;
    inorder(root);
    for(int i=0;i<in.size();i++){
        cout<<in[i];
        if(i!=in.size()-1)cout<<" ";
    }
    return 0;
}

//我应该是个智障吧，中序递归遍历直接存到一个向量里，最后在输出，不就好了？不直接在便利的时候输出啊！！。。学习了！