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 = 0) :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 - 1, ordL, k - 1);
    //递归构造右子树
    root->r = createTree(posL + numL, posR - 1, k + 1, 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 = 0; i < N; ++i)
        cin >> pos[i];
    for (int i = 0; i < N; ++i)
        cin >> ord[i];
    Node* root = createTree(0, N - 1, 0, N - 1);
    getResBFS(root);
    return 0;    
}