PAT甲 1020 Tree Traversals （树的后序中序->层序）

1020 Tree Traversals (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

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

题意：根据树的后序中序求树的层序
请先看树的后序中序->前序https://www.cnblogs.com/1013star/p/11569194.html

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 using namespace std;
 5 int post[35],in[35];
 6 int a[11000];
 7 //root是后序中的当前根的位置，st,ed是该子树在中序遍历中的最左位置和最右位置 
 8 void dfs(int root,int st,int ed,int index)
 9 {
10     
11     if(st>ed)
12         return;
13     int i=st;
14     while(in[i]!=post[root])
15         i++;
16     //cout<<post[root]<<" ";
17 //    cout<<"index="<<index<<" "<<"root="<<post[root]<<endl;
18     a[index]=post[root];
19     dfs(root-(ed-i+1),st,i-1,index*2);
20     dfs(root-1,i+1,ed,index*2+1);
21  } 
22 int main()
23 {
24     int n;
25     cin>>n;
26     for(int i=0;i<n;i++)
27         cin>>post[i];
28     for(int i=0;i<n;i++)
29         cin>>in[i];
30     dfs(n-1,0,n-1,1);
31     int cnt=0;
32     for(int i=1;i<10000;i++)
33     {
34         if(a[i]!=0)
35         {
36             cnt++;
37             if(cnt==n)
38             {
39                 cout<<a[i];
40                 break;
41             }
42             else
43                 cout<<a[i]<<" "; 
44         }
45     }
46     return 0;    
47 }

posted @ 2019-09-22 21:16 *starry* 阅读(156) 评论(0) 编辑收藏举报

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