PAT 甲级 1102 Invert a Binary Tree(二叉树翻转)

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：题目给出的输入数据是每个节点的左右孩子信息，即节点0的左右孩子分别为节点1与空，节点1没有左右孩子，节点2左右孩子分别为节点0与空，以此类推。此题给出了左右节点的信息，所以用静态链表做比较方便。对二叉树进行反转不需要真正反转，只要将中序遍历的顺序变为（右中左即可），层序遍历每次先将右孩子进队。对于根节点寻找方法，输入数据中没有的数字就是根节点。题目要求结尾没有空格，所以用flag标志来控制输出，只有第一个数据是直接输出，其他都是先输出空格再输出数据。

#include<cstdio>
#include<queue>
#include<string>
#include<iostream>
using namespace std;
struct node{
	int data;
	int lchild;
	int rchild;
}Node[20];
int flag = 0;
void layerorder(int root){   //层序遍历 
	queue<int> q;
	q.push(root);
	while(!q.empty()){
		int front = q.front();
		q.pop();
		if(flag)
			printf(" ");
		printf("%d",Node[front].data);
		flag = 1;
		if(Node[front].rchild!=-1)	q.push(Node[front].rchild);  //先右子树 
		if(Node[front].lchild!=-1)	q.push(Node[front].lchild);  //后左子树 
	}
}

void inorder(int root){
	if(root == -1)
		return;
	inorder(Node[root].rchild);
	if(flag)
		printf(" ");
	printf("%d",Node[root].data);
	flag = 1;
	inorder(Node[root].lchild);
}
int hashtable[20] = {0};
int main(void){
	int N;
	scanf("%d",&N);
	string temp;
	for(int i = 0;i < N;i++){
		Node[i].data = i;
		cin>>temp;
		if(temp[0]>='0'&&temp[0]<='9'){
			Node[i].lchild = temp[0] - '0';
			hashtable[temp[0] - '0'] = 1;
		}
		else
			Node[i].lchild = -1;
		cin>>temp;
		if(temp[0]>='0'&&temp[0]<='9'){
			Node[i].rchild = temp[0] - '0';
			hashtable[temp[0] - '0'] = 1;
		}
		else
			Node[i].rchild = -1;
	}
	int root;
	for(int i = 0;i < N;i++){    //寻找根节点 
		if(hashtable[i]==0){
			root = i;
			break;
		}
	}
	layerorder(root);
	printf("\n");
	flag = 0;
	inorder(root);
	return 0;	
}