PAT甲级——A1020 Tree Traversals

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


已知后序遍历和中序遍历输出层序遍历

 1 #include <iostream>
 2 #include <queue>
 3 using namespace std;
 4 int *pos, *ord;//存放后序和中序遍历数据
 5 struct Node
 6 {
 7     int val;
 8     Node *l, *r;
 9     Node(int a = 0) :val(a), l(nullptr), r(nullptr) {}
10 };
11 Node*  createTree(int posL,int posR, int ordL, int ordR)
12 {
13     if (posL > posR)
14         return nullptr;
15     Node *root = new Node();
16     root->val = pos[posR];//根节点值
17     int k;
18     for (k = ordL; k <= ordR; ++k)
19     {
20         if (ord[k] == pos[posR])//找到原树的根
21             break;
22     }
23     int numL = k - ordL;//左子树节点数量
24     //递归构造左子树
25     root->l = createTree(posL, posL + numL - 1, ordL, k - 1);
26     //递归构造右子树
27     root->r = createTree(posL + numL, posR - 1, k + 1, ordR);//取出根节点
28     return root;
29 }
30 void getResBFS(Node* root)
31 {
32     queue<Node*>q;
33     Node* p = nullptr;
34     q.push(root);
35     cout << root->val;
36     while (!q.empty())
37     {
38         p = q.front();
39         if (p != root)
40             cout << " " << p->val;
41         q.pop();
42         if (p->l != nullptr)
43             q.push(p->l);
44         if (p->r != nullptr)
45             q.push(p->r);
46     }
47     cout << endl;
48 }
49 
50 int main()
51 {
52     int N;
53     cin >> N;
54     pos = new int[N];
55     ord = new int[N];
56     for (int i = 0; i < N; ++i)
57         cin >> pos[i];
58     for (int i = 0; i < N; ++i)
59         cin >> ord[i];
60     Node* root = createTree(0, N - 1, 0, N - 1);
61     getResBFS(root);
62     return 0;    
63 }