PAT甲级--1102 Invert a Binary Tree（25 分）【反转二叉树+层序遍历+中序遍历】

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

解题思路：这里中序遍历和层序遍历都基本上没什么变化，主要是怎么存储这个树节点，怎么找出根节点，这里用到结构，来存储左右孩子节点，当输入的是‘-‘将其赋值为-1，如果是数字就转化存入数字，我们还要找出根节点，只要它不是其他孩子的节点，也就是在输入的这里没有出现过就是根节点。层序遍历和中序遍历我就不多说啦。

 

#include
using namespace std;
struct node{
	int left,right;
};
vectorno;
int flag=0;//控制输出格式
void inOrder(int root)//中序遍历
{
	if(no[root].left==-1&&no[root].right==-1)
	{
		printf("%s%d",flag==0?"":" ",root);
		flag=1;//我这里忘记flag赋值为1，导致一个测试点格式错误
		return;
	}
	if(no[root].left!=-1) inOrder(no[root].left);
	printf("%s%d",flag==0?"":" ",root);
	flag=1; 
	if(no[root].right!=-1) inOrder(no[root].right); 
}
int main(void)
{
	int n;
	scanf("%d",&n);getchar();//忘了getchar（）
	no.resize(n);//要规定大小，才可以用下标输入数据
	vectormark(n);//要规定大小，要不然值输不进去 ，这个用来找出根节点
	for(int i=0;i q;
	q.push(root);
	vectorlevel;
	while(!q.empty())//层序遍历
	{
		int i=q.front();
		q.pop();
		if(no[i].left!=-1)
		q.push(no[i].left);
		if(no[i].right!=-1)
		q.push(no[i].right);
		level.push_back(i);
	}
	int flag1=0;
	for(int i = 0; i < n; i++)
	{
		printf("%s%d", flag1==0?"":" ",level[i]);
		flag1=1;
	}
        cout<