pat甲级1020中序后序求层序

1020 Tree Traversals (25)（25 分）

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=30), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

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

Sample Output:

4 1 6 3 5 7 2

已知后序中序求先序：

 void pre(int root, int begin, int end)
 {
     if (begin > end) return;
         cout  << post[root] << " ";
     int p;
     for (p = begin; p <= end; p++)
     {
         if (in[p] == post[root])
             break;
     }
     pre(root - end + p - , begin, p - );
     pre(root - , p + , end);
 }

后序中最后一个元素为跟，找出跟在中序中的位置就可以确定左右子树的节点个数，然后就可以递归的去求解。

这个题是求层序遍历，开始使用的构造树然后再广度优先搜索的方法，比较繁琐。后来看柳婼的博客发现使用数组的方法构造树比较方便。

若当前根的下标为i，则左节点下标为 2 * i + 1，右节点下标为 2 * i + 2。

 #include <iostream>
 #include <vector>
 #include <math.h>
 using namespace std;

 vector<int> post, in, level(, -);
 void getLevel(int root, int begin, int end, int index);

 int main()
 {
     int N, i;
     cin >> N;
     post.resize(N);
     in.resize(N);
     for (i = ; i < N; i++)
         cin >> post[i];
     for (i = ; i < N; i++)
         cin >> in[i];
     getLevel(N - , , N - , );
     int cnt = ;
     for (int i : level)
     {
         if (i != -)
         {
             if (cnt > )
                 cout << " ";
             cnt++;
             cout << i;
             if (cnt == N)
                 break;
         }
     }
     return ;
 }

 void getLevel(int root, int begin, int end, int index)
 {
     if (begin > end) return;
     level[index] = post[root];
     int p;
     for (p = begin; p <= end; p++)
     {
         if (in[p] == post[root])
             break;
     }
     getLevel(root - end + p - , begin, p - ,  * index + );
     getLevel(root - , p + , end,  * index + );
 }