PAT甲级——A1102 Invert a Binary Tree

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 (≤) 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

思路：
　　先找到根节点，数字r未出现，则r为根节点，因为根节点不是任何节点的子节点
　　然后静态构造树
　　再调用平常的层序遍历和中序遍历
　　所谓的二叉树反转，就是原来先读左子树，再读右子树
　　现在改为先读右子树，再读左子树

 #include <iostream>
 #include <vector>
 #include <queue>
 using namespace std;
 struct Node
 {
     int l, r;
 };
 int N, root[] = {  };
 Node tree[];
 vector<int>lev, in;
 void levelOrde(int t)
 {
     if (t == -)
         return;
     queue<int>q;
     q.push(t);
     while (!q.empty())
     {
         t = q.front();
         q.pop();
         lev.push_back(t);
         if (tree[t].r != -)//先进右
             q.push(tree[t].r);
         if (tree[t].l != -)
             q.push(tree[t].l);
     }
 }
 void inOrder(int t)
 {
     if (t == -)
         return;
     inOrder(tree[t].r);
     in.push_back(t);
     inOrder(tree[t].l);
 }
 int main()
 {
     cin >> N;
     char l, r;
     for (int i = ; i < N; ++i)
     {
         cin >> l >> r;
         if (l != '-')
         {
             tree[i].l = l - '';
             root[l - ''] = -;//去除为根的可能性
         }
         else
             tree[i].l = -;
         if (r != '-')
         {
             tree[i].r = r - '';
             root[r - ''] = -;//去除为根的可能性
         }
         else
             tree[i].r = -;
     }
     for (int i = ; i < N; ++i)
     {
         if (root[i] == )
         {
             r = i;
             break;//找到了根节点
         }
     }
     levelOrde(r);
     inOrder(r);
     for (int i = ; i < N; ++i)
         cout << lev[i] << (i == N -  ? "" : " ");
     cout << endl;
     for (int i = ; i < N; ++i)
         cout << in[i] << (i == N -  ? "" : " ");
     return ;
 }