1102. Invert a Binary Tree (25)<反转二叉树>

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

给一个二叉树，左右结点互换，输出层序遍历和先序遍历

这种类型的题目我比较喜欢用链表做，做的最近这几题，发现直接结构体数组更容易

1099. Build A Binary Search Tree (30) <BST树>

像上面这道题一样，这类题用链表还要开辟空间。一不小心就会出现段错误

建议还是使用数组做

两种方法我也都做了

结构体数组：

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
typedef struct tree{
//int num;
int left,right;
}tree,linktree;
linktree t[100];
int father[100];
int n;
int cengxu[100];
int xianxu[100];
int cnt;
int find(int x){
if(x==father[x]) return x;
return father[x]=find(father[x]);
}
void pro(int x){
if(x!=-1){
pro(t[x].left);
xianxu[cnt++]=x;
//cout<<t[x].num<<" ";
pro(t[x].right);
}
}
int main(){
int n;
cin>>n;
for(int i=0;i<n;i++) father[i]=i;
for(int i=0;i<n;i++){
char a,b;
cin>>a>>b;
if(a=='-'){
t[i].right=-1;
}else{
t[i].right=a-'0';
father[a-'0']=i;
}
if(b=='-'){
t[i].left=-1;
}else{
t[i].left=b-'0';
father[b-'0']=i;
}
}
int start=find(0);
queue<int> que;
que.push(start);
cnt=0;
while(que.size()){
int num=que.front();
que.pop();
cengxu[cnt++]=num;
if(t[num].left!=-1) que.push(t[num].left);
if(t[num].right!=-1) que.push(t[num].right);
}
for(int i=0;i<n;i++) {
if(i==0) cout<<cengxu[i];
else cout<<" "<<cengxu[i];
}
cnt=0;
pro(start);

cout<<endl;
for(int i=0;i<n;i++) {
if(i==0) cout<<xianxu[i];
else cout<<" "<<xianxu[i];
}
return 0;
}


链表：
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
typedef struct tree{
int num;
struct tree *left,*right;
}tree,*linktree;
linktree t[100];
int father[100];
int find(int x){
if(x==father[x]) return x;
return father[x]=find(father[x]);
}
int n;
int cengxu[100];
int xianxu[100],c=0;
void pro(linktree head){
if(head){
pro(head->left);
xianxu[c++]=head->num;
pro(head->right);
}
}
int main(){
int v[100]={0};
cin>>n;
for(int i=0;i<n;i++) father[i]=i;
for(int i=0;i<n;i++){
char a,b;
cin>>a>>b;

if(v[i]==0){
t[i]=new tree();
v[i]=1;
}
t[i]->num=i;
if(a=='-'){
t[i]->right=NULL;
}
else{
if(v[a-'0']==0){
v[a-'0']=1;
t[a-'0']=new tree();
}
t[a-'0']->num=a-'0';
t[i]->right=t[a-'0'];
father[a-'0']=i;
}
if(b=='-'){
t[i]->left=NULL;
}else{
if(v[b-'0']==0){
v[b-'0']=1;
t[b-'0']=new tree();
}
t[b-'0']->num=b-'0';
t[i]->left=t[b-'0'];
father[b-'0']=i;
}
}
int start=find(0);
// cout<<start<<endl;
queue<linktree> que;
que.push(t[start]);
int cnt=0;
while(que.size()){
linktree fz=new tree();
fz=que.front();
que.pop();
cengxu[cnt++]=fz->num;
if(fz->left) que.push(fz->left);
if(fz->right) que.push(fz->right);
}
pro(t[start]);
for(int i=0;i<n;i++) {
if(i==0) cout<<cengxu[i];
else cout<<" "<<cengxu[i];
}
cout<<endl;
for(int i=0;i<n;i++) {
if(i==0) cout<<xianxu[i];
else cout<<" "<<xianxu[i];
}
return 0;
}