【PAT甲级】1102 Invert a Binary Tree（25 分）（二叉树的层序与前序遍历）

题目链接
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

题意：反转一颗给定二叉树，输出反转后的二叉树的层序遍历和前序遍历

思路：这里因为是直接给出的左右孩子节点，反转二叉树也就是所有左右节点交换，直接交换即可。

层序遍历使用queue，bfs。 前序遍历递归，dfs

代码：

#include 
using namespace std;
#define rep(i,a,n) for(int i=a;i=a;i--)
#define mem(a,n) memset(a,n,sizeof(a))
#define lowbit(i) ((i)&(-i))
typedef long long ll;
typedef unsigned long long ull;
const ll INF=0x3f3f3f3f;
const double eps = 1e-6;
const int N = 1e1+5;

struct Node {
    int left,right;
};
vectorlevel,in;
vectortree;
bool vis[N];
int root,n;
void printLevel() {
    for(int i=0; ique;
    que.push(root);
    while(!que.empty()) {
        int node=que.front();
        level.push_back(node);
        que.pop();
        if(tree[node].left!=-1) {
            que.push(tree[node].left);
        }
        if(tree[node].right!=-1) {
            que.push(tree[node].right);
        }
    }
    for(int i=0; i