1102. Invert a Binary Tree (25)

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1102. Invert a Binary Tree (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

N个结点（编号0~N-1；N不超过10）

接着N行分别为编号0~N-1对应点的   右结点编号 左结点编号（如果没有结点就是-）

要求输出这个二叉树的层次遍历的  、中序遍历的

这是这次pat2015中25分的，也是这次pat考试中唯一一个ac并且一次就ac的

评测结果
  时间 结果 得分 题目 语言 用时(ms) 内存(kB) 用户
9月12日
 20:43 答案正确 25 1102 C++
 (g++ 4.7.2) 1 308 datrilla
测试点
  测试点 结果 用时(ms) 内存(kB) 得分/满分
0 答案正确 1 308 13/13
1 答案正确 1 180 5/5
2 答案正确 1 180 5/5
3 答案正确 1 308 2/2
#include<iostream>
#include<vector>
#include<queue> 
using namespace std; 
struct trees
{
  int child[2];
  void functrees(int ll, int rr)
  {
    child[0] = ll;
    child[1] = rr;
  }
};
void level_order(vector<trees>*erCharShu,int root)
{
  queue<int>q;
  int size;
  q.push(root);
  cout << root;
  while (!q.empty())
  {
    size = q.size();
    while (size--)
    {
      root = q.front();
      q.pop();
      if ((*erCharShu)[root].child[0]!= -1)
      {
        cout << " " << (*erCharShu)[root].child[0];
        q.push((*erCharShu)[root].child[0]);
      }
      if ((*erCharShu)[root].child[1] != -1)
      {
        cout << " " << (*erCharShu)[root].child[1];
        q.push((*erCharShu)[root].child[1]);
      }
    }
  }
  cout << endl;
}
void  in_order(vector<trees>*erCharShu,int root,int *alltag)
{
  if ((*erCharShu)[root].child[0] != -1)
    in_order(erCharShu, (*erCharShu)[root].child[0], alltag);
  if (1 == (*alltag))
  { 
    cout << root; 
    (*alltag)++;
  }
  else cout << " " << root;
  if ((*erCharShu)[root].child[1] != -1)
    in_order(erCharShu, (*erCharShu)[root].child[1], alltag);
}
int main()
{
  int N,i,ll,rr;
  cin >> N;
  char r,l;
  vector<bool>flag(N, false);
  vector<trees>erCharShu(N);
  for (i = 0; i < N; i++)
  {
    cin >> r >> l;
    if (r == '-')rr = -1; else
    {
      rr = r -'0';
      flag[rr] = true;
    }
    if (l == '-')ll = -1; else
    {
      ll = l  -'0';
      flag[ll] = true;
    }
    erCharShu[i].functrees(ll,rr);
  }
  for (i = 0; i < N&&flag[i]; i++);
  level_order(&erCharShu, i);
  ll = 1;
  in_order(&erCharShu, i, &ll);
  cout << endl;
  system("pause");
  return 0;
}