PAT_A1020#Tree Traversals

Source:

PAT A1020 Tree Traversals (25 分)

Description:

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 (≤), 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

Keys:

• 二叉树的建立和遍历

Code:

 1 /*
 2 time: 2019-06-30 14:40:45
 3 problem: PAT_A1020#Tree Traversals
 4 AC: 08:33
 5
 6 题目大意：
 7 给出后序和中序遍历，打印层序遍历
 8 */
 9 #include<cstdio>
10 #include<queue>
11 using namespace std;
12 const int M=35;
13 int post[M],in[M],n;
14 struct node
15 {
16     int data;
17     node *lchild,*rchild;
18 };
19
20 node *Create(int postL, int postR, int inL, int inR)
21 {
22     if(postL > postR)
23         return NULL;
24     node *root = new node;
25     root->data = post[postR];
26     int k;
27     for(k=inL; k<=inR; k++)
28         if(in[k] == post[postR])
29             break;
30     int numLeft = k-inL;
31     root->lchild = Create(postL,postL+numLeft-1,inL,k-1);
32     root->rchild = Create(postL+numLeft,postR-1,k+1,inR);
33     return root;
34 }
35
36 void LayerOrder(node *root)
37 {
38     queue<node*> q;
39     q.push(root);
40     int pt=0;
41     while(!q.empty())
42     {
43         root = q.front();
44         q.pop();
45         printf("%d%c", root->data, ++pt==n?‘\n‘:‘ ‘);
46         if(root->lchild)
47             q.push(root->lchild);
48         if(root->rchild)
49             q.push(root->rchild);
50     }
51 }
52
53 int main()
54 {
55 #ifdef ONLINE_JUDGE
56 #else
57     freopen("Test.txt", "r", stdin);
58 #endif // ONLINE_JUDGE
59
60     scanf("%d", &n);
61     for(int i=0; i<n; i++)
62         scanf("%d", &post[i]);
63     for(int i=0; i<n; i++)
64         scanf("%d", &in[i]);
65     node *root = Create(0,n-1,0,n-1);
66     LayerOrder(root);
67
68     return 0;
69 }

原文地址：https://www.cnblogs.com/blue-lin/p/11109732.html