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

Tree Traversal Postorder Left Right Root

📖 Scenario: You are working with a simple family tree where each person can have up to two children. You want to visit all family members in a special order: first the left child, then the right child, and finally the parent. This order is called postorder traversal.

🎯 Goal: Build a program that creates a small binary tree, sets up a function to traverse it in postorder (left, right, root), and prints the values in the correct order.

📋 What You'll Learn

Create a binary tree with exactly 3 nodes: root with value 1, left child with value 2, right child with value 3

Create a function called postorderTraversal that visits nodes in left-right-root order

Use recursion inside postorderTraversal to visit left child, then right child, then print the current node's value

Print the values of the nodes in the order they are visited by postorderTraversal

💡 Why This Matters

🌍 Real World

Postorder traversal is used in file system operations, expression tree evaluations, and cleanup tasks where children must be handled before parents.

💼 Career

Understanding tree traversals is essential for software engineers working with hierarchical data, compilers, databases, and many algorithms.

1

Create the binary tree nodes

Create a struct called Node with an int value and two pointers called left and right. Then create three nodes: root with value 1, root->left with value 2, and root->right with value 3.

DSA C++

// Define struct Node and create root, root->left, root->right nodes
// Your code here

Hint

Start by defining the Node structure with value and pointers to left and right children. Then create the root node and assign its left and right children.

2

Create the postorder traversal function

Write a function called postorderTraversal that takes a Node* parameter called node. Inside the function, first check if node is nullptr. If it is, return immediately. Otherwise, call postorderTraversal recursively on node->left, then on node->right.

DSA C++

void postorderTraversal(Node* node) {
    // Your code here
}

Hint

Use recursion to visit the left child first, then the right child. Remember to stop if the node is nullptr.

3

Print the node value after visiting children

Inside the postorderTraversal function, after the recursive calls to postorderTraversal(node->left) and postorderTraversal(node->right), add a line to print the current node's value using std::cout followed by a space.

DSA C++

void postorderTraversal(Node* node) {
    if (node == nullptr) {
        return;
    }
    postorderTraversal(node->left);
    postorderTraversal(node->right);
    // Your code here
}

Hint

After visiting left and right children, print the current node's value with a space to separate values.

4

Call the traversal and print the result

In the main function, call postorderTraversal(root) to start the traversal from the root. Then print a newline using std::cout.

DSA C++

int main() {
    // Call postorderTraversal with root
    // Print a newline
    // Your code here
    return 0;
}

Hint

Call the traversal starting from root and print a newline after to see the output clearly.