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 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
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 }