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 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

#include<iostream>
#include<string>
#include<vector> 
#include<cstdio>
#include<queue>
using namespace std;
struct node{
	int lchild,rchild;
};

struct node a[500];
vector<int> v1,v2;
void level(int root){
	queue<int> q;
	q.push(root);
	while(!q.empty()){
		int x = q.front();
		v1.push_back(x);
		q.pop();
		if(a[x].rchild != -1){
			q.push(a[x].rchild);	
		}
		if(a[x].lchild != -1){
			q.push(a[x].lchild);
		}
	}
}
void dfs(int index){
	if(index == -1)
		return ;
	dfs(a[index].rchild);
	v2.push_back(index);
	dfs(a[index].lchild);
}
int main(){
	int n;
	vector<int> book(500,0);
	cin >> n;
	for(int i = 0;i < n;i++){
		string l,r;
		cin >> l >> r;
		if(l=="-"){
			a[i].lchild = -1;
		}else{
			a[i].lchild = stoi(l);
			book[stoi(l)] = 1;
		}
		if(r=="-"){
			a[i].rchild = -1;
		}else{
			a[i].rchild = stoi(r);
			book[stoi(r)] = 1;
		}
	} 
	int root=0;
	while(book[root] != 0){
		root++;
	}
	level(root);
	for(int i = 0;i < v1.size();i++){
		cout << v1[i];
		if(i != v1.size()-1) cout << " ";
	}
	cout << endl;
	dfs(root);
	for(int i = 0;i < v2.size();i++){
		cout << v2[i];
		if(i != v2.size()-1) cout << " ";
	}
	cout << endl;
	return 0;
}