Why Tree Traversal Postorder Left Right Root in DSA C++?

The Big Idea

Discover how visiting children before parents can save you from costly mistakes in tree operations!

The Scenario

Imagine you have a family tree drawn on paper, and you want to write down the names starting from the youngest children, then their parents, and finally the grandparents. Doing this by hand means you have to carefully check each branch and remember where you left off.

The Problem

Manually visiting each family member in the correct order is slow and easy to mess up. You might forget a child or write a parent before their children, causing confusion and mistakes.

The Solution

Postorder tree traversal automates this process by visiting the left child, then the right child, and finally the parent node. This way, you always process children before their parents, ensuring the correct order without forgetting anyone.

Before vs After

✗ Before

void printPostorder(Node* node) {
  if (node == nullptr) return;
  printPostorder(node->left);
  printPostorder(node->right);
  std::cout << node->value << " ";
}

✓ After

void postorderTraversal(Node* root) {
  if (root == nullptr) return;
  postorderTraversal(root->left);
  postorderTraversal(root->right);
  std::cout << root->value << " ";
}

What It Enables

This traversal lets you process or delete nodes only after all their children are handled, enabling safe and logical tree operations.

Real Life Example

When deleting files in a folder, you must delete all files and subfolders first before deleting the main folder. Postorder traversal mimics this by visiting children before parents.

Key Takeaways

Postorder visits left child, then right child, then root.

It ensures children are processed before their parent.

Useful for tasks like safe deletion or expression evaluation.

Practice

1. What is the order of nodes visited in a postorder traversal of a binary tree?

easy

A. Left subtree, Right subtree, Root node

B. Root node, Left subtree, Right subtree

C. Root node, Right subtree, Left subtree

D. Right subtree, Left subtree, Root node

Solution

1. Step 1: Understand postorder traversal definition

   Postorder traversal visits nodes in the order: left subtree first, then right subtree, and finally the root node.

2. Step 2: Match the traversal order to options

   Left subtree, Right subtree, Root node correctly lists the order as Left, Right, Root which matches postorder traversal.

3. Final Answer:

   Left subtree, Right subtree, Root node -> Option A

4. Quick Check:

   Postorder = Left, Right, Root [OK]

Hint: Remember: postorder ends with root after children [OK]

Common Mistakes:

• Confusing postorder with preorder or inorder
• Swapping left and right subtree order
• Thinking root is visited first

2. Which of the following C++ function signatures correctly represents a postorder traversal of a binary tree node Node*?

easy

A. void postorder(Node* root) { visit(root); postorder(root->left); postorder(root->right); }

B. void postorder(Node* root) { postorder(root->right); postorder(root->left); visit(root); }

C. void postorder(Node* root) { postorder(root->right); visit(root); postorder(root->left); }

D. void postorder(Node* root) { postorder(root->left); postorder(root->right); visit(root); }

Solution

1. Step 1: Recall postorder traversal steps

   Postorder visits left subtree, then right subtree, then root node.

2. Step 2: Check function calls order

   void postorder(Node* root) { postorder(root->left); postorder(root->right); visit(root); } calls postorder on left, then right, then visits root, matching postorder traversal.

3. Final Answer:

   void postorder(Node* root) { postorder(root->left); postorder(root->right); visit(root); } -> Option D

4. Quick Check:

   Left, Right, Root call order = void postorder(Node* root) { postorder(root->left); postorder(root->right); visit(root); } [OK]

Hint: Postorder calls left, then right, then root visit [OK]

Common Mistakes:

• Visiting root before children (preorder)
• Swapping left and right subtree calls
• Missing visit call after recursion

3. Given the binary tree:

    1
   / \
  2   3
 / \
4   5

What is the output of a postorder traversal?

medium

A. [1, 2, 3, 4, 5]

B. [4, 2, 5, 3, 1]

C. [4, 5, 2, 3, 1]

D. [2, 4, 5, 3, 1]

Solution

1. Step 1: Traverse left subtree of root (1)

   Left subtree rooted at 2: postorder visits 4, then 5, then 2.

2. Step 2: Traverse right subtree and root

   Right subtree is 3 alone, visited next. Finally, root 1 is visited.

3. Final Answer:

   [4, 5, 2, 3, 1] -> Option C

4. Quick Check:

   Postorder = Left, Right, Root = [4,5,2,3,1] [OK]

Hint: Visit children fully before root [OK]

Common Mistakes:

• Listing root before children
• Mixing left and right subtree order
• Confusing with preorder or inorder output

4. The following C++ code attempts postorder traversal but produces incorrect output. What is the error?

void postorder(Node* root) {
  if (!root) return;
  postorder(root->right);
  postorder(root->left);
  visit(root);
}

medium

A. Left and right subtree calls are swapped

B. The base case check is incorrect

C. The visit call is missing

D. The function should visit root before children

Solution

1. Step 1: Analyze recursive calls order

   The code calls postorder on right subtree first, then left subtree, which reverses the correct order.

2. Step 2: Understand correct postorder order

   Postorder requires left subtree first, then right subtree, then root visit.

3. Final Answer:

   Left and right subtree calls are swapped -> Option A

4. Quick Check:

   Left before right in postorder [OK]

Hint: Left subtree call must come before right [OK]

Common Mistakes:

• Swapping left and right subtree calls
• Visiting root before children
• Omitting base case check

5. You want to collect all node values of a binary tree in postorder traversal into a vector. Which code snippet correctly implements this in C++?

hard

A. void postorder(Node* root, vector<int>& res) { if (!root) return; res.push_back(root->val); postorder(root->left, res); postorder(root->right, res); }

B. void postorder(Node* root, vector<int>& res) { if (!root) return; postorder(root->left, res); postorder(root->right, res); res.push_back(root->val); }

C. void postorder(Node* root, vector<int>& res) { if (!root) return; postorder(root->right, res); postorder(root->left, res); res.push_back(root->val); }

D. void postorder(Node* root, vector<int>& res) { if (!root) return; postorder(root->left, res); res.push_back(root->val); postorder(root->right, res); }

Solution

1. Step 1: Confirm traversal order in code

   void postorder(Node* root, vector<int>& res) { if (!root) return; postorder(root->left, res); postorder(root->right, res); res.push_back(root->val); } calls postorder on left, then right subtree, then adds root value to vector, matching postorder.

2. Step 2: Check other options for order mistakes

   The other options either add the root value too early or swap left and right calls, breaking postorder.

3. Final Answer:

   void postorder(Node* root, vector<int>& res) { if (!root) return; postorder(root->left, res); postorder(root->right, res); res.push_back(root->val); } -> Option B

4. Quick Check:

   Left, Right, Root push_back order = void postorder(Node* root, vector<int>& res) { if (!root) return; postorder(root->left, res); postorder(root->right, res); res.push_back(root->val); } [OK]

Hint: Push root value after left and right recursion [OK]

Common Mistakes:

• Pushing root value before children
• Swapping left and right subtree calls
• Forgetting to check for null root