【PAT甲级】1020 Tree Traversals

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🔑中文翻译：树的遍历
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1020 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 (≤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

第一行包含整数 N N N，表示二叉树的节点数。

第二行包含 N N N 个整数，表示二叉树的后序遍历。

第三行包含 N N N 个整数，表示二叉树的中序遍历。

输出一行 N N N 个整数，表示二叉树的层序遍历。

思路

1. 由于题目给定每个结点的值都不同，所以可以用哈希表存储每个结点在中序数组中的下标，方便后续使用。

2. 通过下标构建二叉树（其中 k-1-il 表示左子树结点个数）：

3. 通过 bfs 进行层次遍历，并输出最终结果。

代码

#include
using namespace std;

const int N = 40;
int n;
int posorder[N], inorder[N];
unordered_map<int, int> l, r, pos;
int q[N];

//构建树
int build(int il, int ir, int pl, int pr)
{
    int root = posorder[pr];  //获得根结点
    int k = pos[root];    //获得该结点在中序数组中的下标
    if (il < k)   l[root] = build(il, k - 1, pl, pl + (k - 1 - il));        //递归左子树
    if (k < ir)   r[root] = build(k + 1, ir, pl + (k - 1 - il) + 1, pr - 1);    //递归右子树
    return root;
}

//层次遍历
void bfs(int root)
{
    //初始化，用队列来模拟层次遍历
    int hh = 0, tt = 0;
    q[0] = root;

    //开始遍历
    while (hh <= tt)
    {
        int k = q[hh++];
        if (l.count(k))  q[++tt] = l[k];
        if (r.count(k))  q[++tt] = r[k];
    }

    //输出结果
    cout << q[0];
    for (int i = 1; i < n; i++)    cout << " " << q[i];
    cout << endl;
}

int main()
{
    cin >> n;
    for (int i = 0; i < n; i++)    cin >> posorder[i];
    for (int i = 0; i < n; i++)
    {
        cin >> inorder[i];
        pos[inorder[i]] = i;  //记录该值在中序数组中的下标
    }

    int root = build(0, n - 1, 0, n - 1);

    bfs(root);

    return 0;
}