PAT-PAT (Advanced Level) Practise 1102 Invert a Binary Tree (25)【三星级】

题目链接：http://www.patest.cn/contests/pat-a-practise/1102

题面：

1102. Invert a Binary Tree (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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：

    给定一棵树，将其反转后，输出层序遍历和中序遍历的结果。

解题：

    反转操作很简单，建树的时候，交换左右节点即可。层序遍历用bfs，中序遍历用dfs。一开始想为什么数据只有10，这么小。其实这也是一个坑点，如果大于10，那么读入的时候就肯定坑很多人。

代码：

#include 
#include 
#include 
#include 
#include 
#define arraysize 100010
using namespace std;
struct node
{
   int le,ri,id;
}nd[12];
int lo[12],io[12];
bool vis[12];
queue  qe;
void level_t(int r)
{
	int cnt=0,tmp;
    qe.push(r);
	while(!qe.empty())
	{
      tmp=qe.front();
	  qe.pop();
	  lo[cnt++]=tmp;
      if(nd[tmp].le>=0)
		  qe.push(nd[tmp].le);
	  if(nd[tmp].ri>=0)
		  qe.push(nd[tmp].ri);
	}
}
int p=0;
void inorder_t(int r)
{
   if(nd[r].le>=0)
	   inorder_t(nd[r].le);
   io[p++]=r;
   if(nd[r].ri>=0)
	   inorder_t(nd[r].ri);
}
int main()
{
  int n,root;
  scanf("%d",&n);
  char a,b;
  memset(vis,0,sizeof(vis));
  for(int i=0;i='0'&&a<='9')
      {
		  nd[i].ri=a-'0';
		  vis[a-'0']=1;
	  }
	  else
		  nd[i].ri=-1;
	  if(b>='0'&&b<='9')
	  {
		  nd[i].le=b-'0';
		  vis[b-'0']=1;
	  }
	  else
		  nd[i].le=-1;
  }
  for(int i=0;i