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=-;right=-;father=-;
}
}node[];
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!=-)
inorder(node[r].left);
cout<<r<<" ";
if(node[r].right!=-)
inorder(node[r].right);
}
int main() {
int n;
cin>>n;
char ch1,ch2;
for(int i=;i<n;i++){
cin>>ch1>>ch2;
if(ch1!='-'){
node[i].right=ch1-'';
node[ch1-''].father=i;
}
if(ch2!='-'){
node[i].left=ch2-'';
node[ch2-''].father=i;
}
}
root=-;
for(int i=;i<n;i++){
if(node[i].father==-){
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!=-)que.push(node[top].left);
if(node[top].right!=-)que.push(node[top].right);
if(!que.empty())cout<<" ";
}
cout<<endl;
inorder(root); return ;
}
#include <iostream>
#include <map>
#include <cstdio>
#include <queue>
using namespace std;
struct Node{
int father;
int left,right;
Node(){
left=-;right=-;father=-;
}
}node[];
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!=-)
inorder(node[r].left);
in.push_back(r);
if(node[r].right!=-)
inorder(node[r].right);
}
int main() {
int n;
cin>>n;
char ch1,ch2;
for(int i=;i<n;i++){
cin>>ch1>>ch2;
if(ch1!='-'){
node[i].right=ch1-'';
node[ch1-''].father=i;
}
if(ch2!='-'){
node[i].left=ch2-'';
node[ch2-''].father=i;
}
}
root=-;
for(int i=;i<n;i++){
if(node[i].father==-){
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!=-)que.push(node[top].left);
if(node[top].right!=-)que.push(node[top].right);
if(!que.empty())cout<<" ";
}
cout<<endl;
inorder(root);
for(int i=;i<in.size();i++){
cout<<in[i];
if(i!=in.size()-)cout<<" ";
}
return ;
}

//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!

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