Tree Traversal Postorder Left Right Root in DSA C - Time & Space Complexity

Time Complexity: Tree Traversal Postorder Left Right Root

O(n)

Understanding Time Complexity

We want to understand how the time needed to visit all nodes in a tree grows as the tree gets bigger.

Specifically, we ask: How long does postorder traversal take as the number of nodes increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void postorderTraversal(Node* root) {
    if (root == NULL) return;
    postorderTraversal(root->left);
    postorderTraversal(root->right);
    printf("%d ", root->data);
}
    

This code visits every node in a binary tree in postorder: left child, then right child, then the node itself.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls visiting each node once.
• How many times: Exactly once per node in the tree.

How Execution Grows With Input

As the number of nodes grows, the function visits each node once, so the work grows directly with the number of nodes.

Input Size (n)	Approx. Operations
10	About 10 visits
100	About 100 visits
1000	About 1000 visits

Pattern observation: The time grows linearly as the tree size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to complete the traversal grows directly in proportion to the number of nodes.

Common Mistake

[X] Wrong: "Postorder traversal takes more than linear time because it visits nodes multiple times."

[OK] Correct: Each node is visited exactly once, even though recursion happens, so the total work is linear.

Interview Connect

Understanding this helps you explain how recursive tree traversals work efficiently, a common topic in interviews.

Self-Check

"What if we changed the traversal to visit nodes in preorder (root, left, right)? How would the time complexity change?"

Practice

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

easy

A. Right subtree, Left subtree, Root node

B. Root node, Left subtree, Right subtree

C. Root node, Right subtree, Left subtree

D. Left subtree, Right subtree, Root node

Solution

1. Step 1: Understand postorder traversal definition

   Postorder traversal visits the left subtree first, then the right subtree, and finally the root node.

2. Step 2: Apply the order to the traversal

   Following this order ensures all children are visited before the parent node.

3. Final Answer:

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

4. Quick Check:

   Postorder = Left, Right, Root [OK]

Hint: Remember: postorder ends with the root node [OK]

Common Mistakes:

• Confusing postorder with preorder or inorder
• Visiting root before children
• Swapping left and right subtree order

2. Which of the following C code snippets correctly implements the postorder traversal function for a binary tree node struct Node?

easy

A. void postorder(struct Node* root) { if(root) { printf("%d ", root->data); postorder(root->left); postorder(root->right); } }

B. void postorder(struct Node* root) { if(root) { postorder(root->left); postorder(root->right); printf("%d ", root->data); } }

C. void postorder(struct Node* root) { if(root) { postorder(root->right); postorder(root->left); printf("%d ", root->data); } }

D. void postorder(struct Node* root) { if(root) { postorder(root->left); printf("%d ", root->data); postorder(root->right); } }

Solution

1. Step 1: Check traversal order in code

   Postorder requires visiting left subtree, then right subtree, then root node.

2. Step 2: Match code to postorder steps

   void postorder(struct Node* root) { if(root) { postorder(root->left); postorder(root->right); printf("%d ", root->data); } } calls postorder on left, then right, then prints root data, matching postorder.

3. Final Answer:

   Correct postorder traversal code -> Option B

4. Quick Check:

   Left, Right, Root in code = void postorder(struct Node* root) { if(root) { postorder(root->left); postorder(root->right); printf("%d ", root->data); } } [OK]

Hint: Print root after left and right recursive calls [OK]

Common Mistakes:

• Printing root before children (preorder)
• Swapping left and right subtree calls
• Missing null check for root

3. Given the binary tree below:

    1
   / \
  2   3
 / \
4   5

What is the output of a postorder traversal?

medium

A. 2 4 5 3 1

B. 1 2 4 5 3

C. 4 5 2 3 1

D. 4 2 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 (leaf), then visit root 1 last.

3. Final Answer:

   4 5 2 3 1 -> Option C

4. Quick Check:

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

Hint: Visit children fully before printing parent [OK]

Common Mistakes:

• Printing root too early
• Mixing left and right subtree order
• Confusing preorder or inorder output

4. Consider the following C function for postorder traversal:

void postorder(struct Node* root) {
    if(root == NULL) return;
    postorder(root->right);
    postorder(root->left);
    printf("%d ", root->data);
}

What is the main issue with this code?

medium

A. It visits the right subtree before the left subtree, reversing the correct order

B. It does not check if root is NULL before recursion

C. It prints the root before visiting children

D. It causes infinite recursion

Solution

1. Step 1: Analyze the order of recursive calls

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

2. Step 2: Check printing position

   Printing root after children is correct, but subtree order is wrong.

3. Final Answer:

   Visits right subtree before left, reversing order -> Option A

4. Quick Check:

   Left before right needed, code does right before left [OK]

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

Common Mistakes:

• Swapping left and right subtree calls
• Printing root too early
• Missing null check (not in this code though)

5. You have a binary tree where each node stores an integer. You want to collect all node values in postorder traversal into an array. Which approach correctly achieves this?

hard

A. Use recursion: traverse left subtree, traverse right subtree, then append root value to array

B. Use recursion: append root value, then traverse left and right subtrees

C. Use a queue to enqueue root first, then left and right children

D. Traverse the tree level by level and append nodes to array

Solution

1. Step 1: Understand postorder traversal for collection

   Postorder means visiting left subtree, right subtree, then root, so values must be appended in that order.

2. Step 2: Choose correct method to collect values

   Recursion that appends root after children ensures correct postorder sequence in the array.

3. Final Answer:

   Recursion: left, right, then append root -> Option A

4. Quick Check:

   Append after children = postorder collection [OK]

Hint: Append root after children for postorder array [OK]

Common Mistakes:

• Appending root before children (preorder)
• Using queue for level order, not postorder
• Confusing traversal types