PAT 甲级 1102. Invert a Binary Tree (25)

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 

#include <algorithm>
#include <cstdio>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
vector<int> in;
struct Tree {
	int left, right;
};
vector<Tree> tree;
void inorder(int root) {
	if (tree[root].left == -1 && tree[root].right == -1) {
		in.push_back(root);
		return;
	}
	if (tree[root].left != -1)
		inorder(tree[root].left);
	in.push_back(root);
	if (tree[root].right != -1)
		inorder(tree[root].right);
}

int main() {
	int n, root;
	scanf("%d", &n);
	getchar();
	tree.resize(n);
	vector<int> book(n);
	for (int i = 0; i < n; i++) {
		char c1, c2;
		scanf("%c %c", &c1, &c2); 
		getchar();
		tree[i].right = (c1 == '-' ? -1 : (c1 - '0'));
		tree[i].left = (c2 == '-' ? -1 : (c2 - '0'));
		if (tree[i].left != -1)
			book[tree[i].left] = 1;
		if (tree[i].right != -1)
			book[tree[i].right] = 1;
	}
	for (int i = 0; i < n; i++) {

		if (book[i] == 0) {
			root = i;
			break;
		}
	}
	queue<int> q;
	q.push(root);
	vector<int> level;
	while (!q.empty()) {
		int node = q.front();
		q.pop();
		if (tree[node].left != -1)
			q.push(tree[node].left);
		if (tree[node].right != -1)
			q.push(tree[node].right);
		level.push_back(node);
	}
	for (int i = 0; i < n; i++) 
		printf("%d%c", level[i], i == n - 1 ? '\n' : ' ');
		inorder(root);
		printf("%d", in[0]);
		for (int i = 1; i < n; i++)
			printf(" %d", in[i]);
		return 0;
}