PAT(树)——1102 Invert a Binary Tree （25 分）

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 *** 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：

给出一棵反转二叉树分别求它的层序遍历和中序遍历序列。

题目解析：

注意数据读入时前面一个是右孩子，后面是左孩子，没有直接给出根节点，只需要记录所有的孩子节点，根节点不是任何结点的孩子。

具体代码：

#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
#define N 15
struct node{
	int lchild=-1,rchild=-1;
}A[N];
bool isC[N]={false};
vector<int> v;
void levelorder(int root){
	queue<int> q;
	q.push(root);
	while(q.size()!=0){
		int t=q.front();q.pop();
		v.push_back(t);
		if(A[t].lchild!=-1)
			q.push(A[t].lchild);
		if(A[t].rchild!=-1)
			q.push(A[t].rchild);
	}
}
void inorder(int root){
	if(root==-1)
		return;
	inorder(A[root].lchild);
	v.push_back(root);
	inorder(A[root].rchild);
}
void print(){
	for(int i=0;i<v.size();i++){
		cout<<v[i];
		if(i!=v.size()-1)
			cout<<" ";
	}
}
int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;i++){
    	char rchild,lchild;
    	getchar();
    	scanf("%c %c",&rchild,&lchild);
    	if(rchild!='-'){
    		A[i].rchild=(int)(rchild-'0');
    		isC[(int)(rchild-'0')]=true;
		}
		if(lchild!='-'){
			A[i].lchild=(int)(lchild-'0');
    		isC[(int)(lchild-'0')]=true;
		}
	}
	int root=-1;
	for(int i=0;i<n;i++)
		if(isC[i]==false)
			root=i;
			
	levelorder(root);
	print();
	cout<<endl;
	v.clear();
	inorder(root);
	print();
    return 0;
}