Complexity for print root to leaf of a perfect binary tree

In, the code for printing root to leaf for every leaf node is provided below.

They state the algorithm is O(n), but I think it should be O(n log n) where n is the number of nodes. A standard DFS is typically O(n + E), but printing the paths seems to add a log n. Suppose $h$ is the height of the perfect binary tree. There are $n/2$ nodes on the last level, hence $n/2$ paths that we need to print. Each path has $h + 1$ (let’s just say it’s $h$ for mathematical simplicity) nodes. So we need end up printing $h * frac{n}{2}$ nodes when printing all the paths. We know $h = log_2(n)$. So $h * frac{n}{2} =

#include <iostream>
#include <vector>
using namespace std;
// Data structure to store a binary tree node
struct Node
    int data;
    Node *left, *right;
    Node(int data)
        this->data = data;
        this->left = this->right = nullptr;
// Function to check if a given node is a leaf node or not
bool isLeaf(Node* node) {
    return (node->left == nullptr && node->right == nullptr);
// Recursive function to find paths from the root node to every leaf node
void printRootToleafPaths(Node* node, vector<int> &path)
    // base case
    if (node == nullptr) {
    // include the current node to the path
    // if a leaf node is found, print the path
    if (isLeaf(node))
        for (int data: path) {
            cout << data << " ";
        cout << endl;
    // recur for the left and right subtree
    printRootToleafPaths(node->left, path);
    printRootToleafPaths(node->right, path);
    // backtrack: remove the current node after the left, and right subtree are done
// The main function to print paths from the root node to every leaf node
void printRootToleafPaths(Node* node)
    // vector to store root-to-leaf path
    vector<int> path;
    printRootToleafPaths(node, path);
int main()
    /* Construct the following tree
          2       3
         /      / 
        4   5   6   7
              8       9
    Node* root = new Node(1);
    root->left = new Node(2);
    root->right = new Node(3);
    root->left->left = new Node(4);
    root->left->right = new Node(5);
    root->right->left = new Node(6);
    root->right->right = new Node(7);
    root->right->left->left = new Node(8);
    root->right->right->right = new Node(9);
    // print all root-to-leaf paths
    return 0;