已知二叉树的先序遍历序列和中序遍历序列，使用C++输出其层次序遍历序列

可以使用递归的方式构建二叉树，并利用队列实现层次遍历。以下是示例代码：

#include <iostream>
#include <queue>
using namespace std;

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder, int preStart, int preEnd, int inStart, int inEnd) {
    if (preStart > preEnd || inStart > inEnd) {
        return NULL;
    }
    int rootVal = preorder[preStart];
    int rootIndex = 0;
    for (int i = inStart; i <= inEnd; i++) {
        if (inorder[i] == rootVal) {
            rootIndex = i;
            break;
        }
    }
    int leftSize = rootIndex - inStart;
    TreeNode* root = new TreeNode(rootVal);
    root->left = buildTree(preorder, inorder, preStart + 1, preStart + leftSize, inStart, rootIndex - 1);
    root->right = buildTree(preorder, inorder, preStart + leftSize + 1, preEnd, rootIndex + 1, inEnd);
    return root;
}

void levelOrder(TreeNode* root) {
    if (root == NULL) {
        return;
    }
    queue<TreeNode*> q;
    q.push(root);
    while (!q.empty()) {
        int size = q.size();
        for (int i = 0; i < size; i++) {
            TreeNode* curr = q.front();
            q.pop();
            cout << curr->val << " ";
            if (curr->left != NULL) {
                q.push(curr->left);
            }
            if (curr->right != NULL) {
                q.push(curr->right);
            }
        }
    }
}

int main() {
    vector<int> preorder = { 3, 9, 20, 15, 7 };
    vector<int> inorder = { 9, 3, 15, 20, 7 };
    TreeNode* root = buildTree(preorder, inorder, 0, preorder.size() - 1, 0, inorder.size() - 1);
    levelOrder(root); // 输出 3 9 20 15 7
    return 0;
}