pat甲级1020中序后序求层序

Posted 2020-12-30 lxc1910

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：

 1 void pre(int root, int begin, int end)
 2 {
 3     if (begin > end) return;
 4         cout  << post[root] << " "; 
 5     int p;
 6     for (p = begin; p <= end; p++)
 7     {
 8         if (in[p] == post[root])
 9             break;
10     }
11     pre(root - end + p - 1, begin, p - 1);
12     pre(root - 1, p + 1, end);
13 }

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。

 1 #include <iostream>
 2 #include <vector>
 3 #include <math.h>
 4 using namespace std;
 5 
 6 vector<int> post, in, level(100000, -1);
 7 void getLevel(int root, int begin, int end, int index);
 8 
 9 int main()
10 {
11     int N, i;
12     cin >> N;
13     post.resize(N);
14     in.resize(N);
15     for (i = 0; i < N; i++)
16         cin >> post[i];
17     for (i = 0; i < N; i++)
18         cin >> in[i];
19     getLevel(N - 1, 0, N - 1, 0);
20     int cnt = 0;
21     for (int i : level)
22     {
23         if (i != -1)
24         {
25             if (cnt > 0)
26                 cout << " ";
27             cnt++;
28             cout << i;
29             if (cnt == N)
30                 break;
31         }
32     }
33     return 0;
34 }
35 
36 void getLevel(int root, int begin, int end, int index)
37 {
38     if (begin > end) return;
39     level[index] = post[root];
40     int p;
41     for (p = begin; p <= end; p++)
42     {
43         if (in[p] == post[root])
44             break;
45     }
46     getLevel(root - end + p - 1, begin, p - 1, 2 * index + 1);
47     getLevel(root - 1, p + 1, end, 2 * index + 2);
48 }