Tree Traversal Preorder Root Left Right in DSA C++ - Build from Scratch

Tree Traversal Preorder Root Left Right

📖 Scenario: Imagine you have a family tree and you want to visit each family member starting from the oldest ancestor, then their children from left to right.

🎯 Goal: You will build a simple binary tree and write code to visit each node in preorder: first the root, then the left child, then the right child.

📋 What You'll Learn

Create a binary tree node structure with integer values

Build a small tree with exactly 3 nodes: root, left child, and right child

Write a preorder traversal function that visits nodes in root-left-right order

Print the values of nodes as they are visited

💡 Why This Matters

🌍 Real World

Tree traversal is used in many areas like file system navigation, organizing data, and searching.

💼 Career

Understanding tree traversal is essential for software development roles involving data structures, algorithms, and system design.

1

Create the Tree Node Structure and Root Node

Create a struct called Node with an int data field and two Node* pointers called left and right. Then create a Node* variable called root and initialize it with a new Node having data = 1, and left and right set to nullptr.

DSA C++

// Your code here

Hint

Think of Node as a box holding a number and two pointers to other boxes.

2

Add Left and Right Children to the Root

Create two new Node objects with data = 2 and data = 3, both with left and right set to nullptr. Assign the node with data = 2 to root->left and the node with data = 3 to root->right.

DSA C++

struct Node {
    int data;
    Node* left;
    Node* right;
};

Node* root = new Node{1, nullptr, nullptr};
// Your code here

Hint

Attach two new boxes to the left and right pointers of the root box.

3

Write the Preorder Traversal Function

Write a function called preorder that takes a Node* parameter called node. If node is not nullptr, first print node->data, then recursively call preorder on node->left, then on node->right.

DSA C++

struct Node {
    int data;
    Node* left;
    Node* right;
};

Node* root = new Node{1, nullptr, nullptr};
root->left = new Node{2, nullptr, nullptr};
root->right = new Node{3, nullptr, nullptr};

// Your code here

Hint

Visit the current node first, then the left child, then the right child.

4

Call Preorder and Print the Result

Call the preorder function with root as the argument. Then print a newline character using cout.

DSA C++

#include 
using namespace std;

struct Node {
    int data;
    Node* left;
    Node* right;
};

Node* root = new Node{1, nullptr, nullptr};
root->left = new Node{2, nullptr, nullptr};
root->right = new Node{3, nullptr, nullptr};

void preorder(Node* node) {
    if (node != nullptr) {
        cout data left);
        preorder(node->right);
    }
}

// Your code here

Hint

Call preorder(root) and print a newline after.