pat甲级1020中序后序求层序

发布时间：2020-12-14 03:49:17 所属栏目：大数据来源：网络整理

导读：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 correspo

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 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 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,p - 1,2 * index + 1);
47     getLevel(root - 1,p + 1,end,2 * index + 2);
48 }

（编辑：李大同）

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