Tree Traversal with 2 threads, printing node data simultaneously - multithreading

Question

We are given two threads which perform a pre-order traversal. We have to print data such that for the following example tree:
1
/ \
2 3
/ \ / \
4 NULL NULL NULL
/ \
NULL NULL
...the output is: 1 1 2 2 4 4 3 3
Here the first "thread1" should print one value, and then "thread2" should print a value.

Answer 1

Following is one of possible solution is there any other optimise solution:
#include <iostream>
#include <bits/stdc++.h>
#include <thread>
#include <mutex>
#include <condition_variable>
using namespace std;
bool ready1 = false;
bool ready2 = false;
mutex m;
condition_variable cond;
typedef struct Node {
int data;
struct Node* left;
struct Node* right;
Node(int val)
{
data = val;
left = NULL;
right = NULL;
}
} Node;
void prod1(Node *root) {
if (root == NULL) {
return;
}
unique_lock<mutex> lock(m);
cond.wait(lock, [](){
return ready1;
});
cout<<std::this_thread::get_id()<<" "<<root->data<<endl;
ready2 = true;
ready1 = false;
lock.unlock();
cond.notify_one();
prod1(root->left);
prod1(root->right);
}
void prod2(Node *root) {
if (root == NULL) {
return;
}
unique_lock<mutex> lock(m);
cond.wait(lock, [](){
return ready2;
});
cout<<std::this_thread::get_id()<<" "<<root->data<<endl;
ready1 = true;
ready2 = false;
lock.unlock();
cond.notify_one();
prod2(root->left);
prod2(root->right);
}
int main()
{
/*create root*/
struct Node* root = new Node(1);
root->left = new Node(2);
root->right = new Node(3);
root->left->left = new Node(4);
/* 4 becomes left child of 2
1
/ \
2 3
/ \ / \
4 NULL NULL NULL
/ \
NULL NULL
*/
thread t1(prod1, root);
thread t2(prod2, root);
{
std::lock_guard<std::mutex> lk(m);
ready1 = true;
// std::cout << "main() signals data ready for processing\n";
}
t1.join();
t2.join();
//preorder(root);
return 0;
}