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.

8
1 -
- -
0 -
2 7
- -
- -
5 -
4 6

3 7 2 6 4 0 5 1
6 5 7 4 3 2 0 1

思路

1.題目要求左右顛倒二叉樹，並按層次遍歷和中序遍歷輸出。那麽其實只要在構造樹的時候交換下輸入數據就可以直接構造出一顆顛倒後的樹了。
2.輸出的時候需要註意空格，對於兩種遍歷的輸出只要特殊標識下第一次的輸出就行了。

代碼

#include<iostream>
#include<vector>
#include<queue>
using namespace std;
class Node
{
public:
  int left;
  int right;
  int value;
};


vector<Node> btree(10);

int createTree(const int& N)
{
   vector<bool> roots(N,true);
    for(int i = 0;i < N ;i++)
    {
      char l,r;
      cin >> l >> r;
      btree[i].value = i;
      //invert left
      if(l != ‘-‘)
      {
        btree[i].right = l - ‘0‘;
        roots[l-‘0‘] = false;
      }
      else
        btree[i].right = -1;
      //invert right
      if(r != ‘-‘)
      {
        btree[i].left = r - ‘0‘;
        roots[r-‘0‘] = false;
      }
      else
        btree[i].left = - 1;
    }
    int root = 0;
    for(int i = 0;i < N;i++)
    {
        if(roots[i] == true)
        {
            root = i;
            break;
        }
    }
    return root;
}

void bfs(int root)
{
  queue<int> q;
  q.push(root);
  while(!q.empty())
  {
      int cur = q.front();
      q.pop();
      if(cur == root)
        cout << cur;
      else
        cout << " " << cur;
      if(btree[cur].left != - 1)
        q.push(btree[cur].left);
      if(btree[cur].right != -1)
        q.push(btree[cur].right);
  }
  cout << endl;
}

int firstput = 0;
void inorder(int root)
{
  if(root == -1)
    return;
  if(btree[root].left != -1)
    inorder(btree[root].left);

  if( firstput++ == 0)
    cout << root;
  else
    cout << " " <<root;

  if(btree[root].right != -1)
    inorder(btree[root].right);
}

int main()
{
  int N;
  while(cin >> N)
  {
     int root = createTree(N);
     bfs(root);

     inorder(root);
  }
}

PAT1102: Invert a Binary Tree