PAT甲级 1020 Tree Traversals (25 分)

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本文介绍了一种根据给定的后序遍历和中序遍历序列构建二叉树，并输出该树的层次遍历序列的方法。通过递归地创建二叉树节点，最后使用队列实现层次遍历。

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

#include <iostream>
#include <queue>

using namespace std;
int PostOrder[30];
int InOrder[30];
struct node {
    node *lchild;
    node *rchild;
    int data;
};

node *create(int postL, int postR, int inL, int inR) {
    if (postR < postL)
        return nullptr;
    node *root = new node;
    root->data = PostOrder[postR];
    int k;
    for (k = inL; k <= inR; ++k) {
        if (PostOrder[postR] == InOrder[k])
            break;
    }
    int numLeft = k - inL;
    root->lchild = create(postL, postL + numLeft - 1, inL, k - 1);
    root->rchild = create(postL + numLeft, postR - 1, k + 1, inR);
    return root;
}

void LayerOrder(node *root) {
    queue<node *> q;
    q.push(root);
    cout << root->data;
    int count = 0;
    while (!q.empty()) {
        node *ans = q.front();
        if (count)
            cout << " " << ans->data;
        q.pop();
        if (ans->lchild != nullptr)
            q.push(ans->lchild);
        if (ans->rchild != nullptr)
            q.push(ans->rchild);
        count++;
    }
}

int main() {
    int N;
    cin >> N;
    for (int i = 0; i < N; ++i) {
        cin >> PostOrder[i];
    }
    for (int i = 0; i < N; ++i) {
        cin >> InOrder[i];
    }
    node *root = create(0, N - 1, 0, N - 1);
    LayerOrder(root);
    return 0;
}