用C++语言求一棵二叉树中每层叶节点的数目，要求： 编写函数 LeafInEachLevel(root)，其中：root是指向二叉树根节点的指针； 该函数的功能是，输出以root为根的二叉树中每层的叶节点数目。

好的，以下是用C++语言实现的代码：

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

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

vector<int> LeafInEachLevel(TreeNode* root) {
    vector<int> res;
    if (!root) return res;
    queue<TreeNode*> q{{root}};
    int level = 0, leafCount = 0, nextLevelCount = 1;
    while (!q.empty()) {
        auto node = q.front(); q.pop();
        if (!node->left && !node->right) ++leafCount;
        if (node->left) q.push(node->left);
        if (node->right) q.push(node->right);
        if (--nextLevelCount == 0) {
            res.push_back(leafCount);
            leafCount = 0;
            ++level;
            nextLevelCount = q.size();
        }
    }
    return res;
}

int main() {
    // 构建二叉树
    TreeNode* root = new TreeNode(1);
    root->left = new TreeNode(2);
    root->right = new TreeNode(3);
    root->left->left = new TreeNode(4);
    root->left->right = new TreeNode(5);
    root->right->left = new TreeNode(6);
    root->right->right = new TreeNode(7);

    vector<int> res = LeafInEachLevel(root);
    for (int i = 0; i < res.size(); ++i) {
        cout << "Level " << i << " leaf count: " << res[i] << endl;
    }
    return 0;
}

这个函数的时间复杂度也是O(n)，其中n是二叉树中节点的个数。