HackerRank•Jun 23, 2025
Preorder Traversal
Hazrat Ali
HackerRank
Complete the function in the editor below, which has parameter: a pointer to the root of a binary tree. It must print the values in the tree's preorder traversal as a single line of space-separated values.
Input Format
Our test code passes the root node of a binary tree to the preOrder function.
Constraints
Nodes in the tree
Output Format
Print the tree's preorder traversal as a single line of space-separated values.
Sample Input
1
\
2
\
5
/ \
3 6
\
4
Sample Output
1 2 5 3 4 6
Solution
#include <bits/stdc++.h>
using namespace std;
class Node {
public:
int data;
Node *left;
Node *right;
Node(int d) {
data = d;
left = NULL;
right = NULL;
}
};
class Solution {
public:
Node* insert(Node* root, int data) {
if(root == NULL) {
return new Node(data);
} else {
Node* cur;
if(data <= root->data) {
cur = insert(root->left, data);
root->left = cur;
} else {
cur = insert(root->right, data);
root->right = cur;
}
return root;
}
}
/* you only have to complete the function given below.
Node is defined as
class Node {
public:
int data;
Node *left;
Node *right;
Node(int d) {
data = d;
left = NULL;
right = NULL;
}
};
*/
void preOrder(Node* root) {
if (!root) return;
cout << root->data << " ";
preOrder(root->left);
preOrder(root->right);
}
}; //End of Solution
int main() {
Solution myTree;
Node* root = NULL;
int t;
int data;
std::cin >> t;
while(t-- > 0) {
std::cin >> data;
root = myTree.insert(root, data);
}
myTree.preOrder(root);
return 0;
}