已知二叉树的先序序列，输出层序遍历序列。

根据二叉树的层序遍历的特点，我们可以使用队列来实现。具体步骤如下：
1. 将先序序列转化为中序序列。
2. 根据中序序列构建二叉树。
3. 将根节点入队。
4. 当队列不为空时，取出队首元素，输出该节点的值，并将其左右子节点入队。
5. 重复步骤4，直到队列为空。

下面是代码示例：

#include <iostream>
#include <queue>
#include <unordered_map>
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, unordered_map<int, int>& map) {
    if (preStart > preEnd) {
        return NULL;
    }
    int rootVal = preorder[preStart];
    int rootIndex = map[rootVal];
    int leftSize = rootIndex - inStart;
    TreeNode* root = new TreeNode(rootVal);
    root->left = buildTree(preorder, inorder, preStart + 1, preStart + leftSize, inStart, rootIndex - 1, map);
    root->right = buildTree(preorder, inorder, preStart + leftSize + 1, preEnd, rootIndex + 1, inEnd, map);
    return root;
}

vector<int> levelOrder(TreeNode* root) {
    vector<int> res;
    if (!root) {
        return res;
    }
    queue<TreeNode*> q;
    q.push(root);
    while (!q.empty()) {
        TreeNode* cur = q.front();
        q.pop();
        res.push_back(cur->val);
        if (cur->left) {
            q.push(cur->left);
        }
        if (cur->right) {
            q.push(cur->right);
        }
    }
    return res;
}

vector<int> preorderToLevelOrder(vector<int>& preorder, vector<int>& inorder) {
    unordered_map<int, int> map;
    for (int i = 0; i < inorder.size(); i++) {
        map[inorder[i]] = i;
    }
    TreeNode* root = buildTree(preorder, inorder, 0, preorder.size() - 1, 0, inorder.size() - 1, map);
    return levelOrder(root);
}

int main() {
    vector<int> preorder = {1, 2, 4, 5, 3, 6, 7};
    vector<int> inorder = {4, 2, 5, 1, 6, 3, 7};
    vector<int> levelorder = preorderToLevelOrder(preorder, inorder);
    for (int i = 0; i < levelorder.size(); i++) {
        cout << levelorder[i] << " ";
    }
    return 0;
}