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

 #include <iostream>
 #include <queue>
 using namespace std;
 int *pos, *ord;//存放后序和中序遍历数据
 struct Node
 {
     int val;
     Node *l, *r;
     Node(int a = ) :val(a), l(nullptr), r(nullptr) {}
 };
 Node*  createTree(int posL,int posR, int ordL, int ordR)
 {
     if (posL > posR)
         return nullptr;
     Node *root = new Node();
     root->val = pos[posR];//根节点值
     int k;
     for (k = ordL; k <= ordR; ++k)
     {
         if (ord[k] == pos[posR])//找到原树的根
             break;
     }
     int numL = k - ordL;//左子树节点数量
     //递归构造左子树
     root->l = createTree(posL, posL + numL - , ordL, k - );
     //递归构造右子树
     root->r = createTree(posL + numL, posR - , k + , ordR);//取出根节点
     return root;
 }
 void getResBFS(Node* root)
 {
     queue<Node*>q;
     Node* p = nullptr;
     q.push(root);
     cout << root->val;
     while (!q.empty())
     {
         p = q.front();
         if (p != root)
             cout << " " << p->val;
         q.pop();
         if (p->l != nullptr)
             q.push(p->l);
         if (p->r != nullptr)
             q.push(p->r);
     }
     cout << endl;
 }

 int main()
 {
     int N;
     cin >> N;
     pos = new int[N];
     ord = new int[N];
     for (int i = ; i < N; ++i)
         cin >> pos[i];
     for (int i = ; i < N; ++i)
         cin >> ord[i];
     Node* root = createTree(, N - , , N - );
     getResBFS(root);
     return ;
 }