Tree Traversal Postorder Left Right Root in DSA Typescript - 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.

function postorderTraversal(node: TreeNode | null): void {
  if (node === null) return;
  postorderTraversal(node.left);
  postorderTraversal(node.right);
  console.log(node.value);
}
    

This code visits every node in the tree by first visiting the left child, then the right child, and finally the node itself.

Identify Repeating Operations

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

How Execution Grows With Input

Each node is visited once, so the total 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 work grows in a straight line with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

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

Common Mistake

[X] Wrong: "Postorder traversal takes longer because it visits nodes multiple times."

[OK] Correct: Each node is visited exactly once, even though the order is left, right, then root. The recursion does not repeat nodes.

Interview Connect

Understanding this traversal's time complexity helps you explain how tree algorithms scale, a key skill in many coding challenges.

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 order of nodes visited in a postorder tree traversal?

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

2. Step 2: Match the traversal order to options

   Left subtree, Right subtree, Root node correctly lists Left subtree, Right subtree, Root node order.

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
• Visiting root before children
• Swapping left and right subtree order

2. Which of the following TypeScript function signatures correctly represents a postorder traversal of a binary tree node?

easy

A. function postorder(node: TreeNode | null): TreeNode[]

B. function postorder(node: TreeNode): number

C. function postorder(): TreeNode

D. function postorder(node: TreeNode | null): void

Solution

1. Step 1: Identify correct parameter and return type for traversal

   Traversal functions usually take a node or null and return void if printing or processing inline.

2. Step 2: Check options for correct signature

   function postorder(node: TreeNode | null): void takes a node or null and returns void, suitable for traversal printing or processing.

3. Final Answer:

   function postorder(node: TreeNode | null): void -> Option D

4. Quick Check:

   Traversal function signature = function postorder(node: TreeNode | null): void [OK]

Hint: Traversal usually takes node|null and returns void [OK]

Common Mistakes:

• Omitting null in parameter type
• Returning incorrect types like TreeNode or number
• Missing parameter in function signature

3. Given the following binary tree:

    1
   / \
  2   3
 / \
4   5

What is the output of a postorder traversal?

medium

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

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

C. [2, 4, 5, 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, visited next. Finally, root 1 is visited.

3. Final Answer:

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

4. Quick Check:

   Postorder output = [4, 5, 2, 3, 1] [OK]

Hint: Postorder prints children before parent [OK]

Common Mistakes:

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

4. Consider this TypeScript postorder traversal function:

function postorder(node: TreeNode | null): void {
  if (!node) return;
  postorder(node.right);
  postorder(node.left);
  console.log(node.val);
}

What is the error in this code?

medium

A. It prints node value before children

B. It does not check for null nodes

C. It visits right subtree before left, reversing postorder

D. It returns a value but should be void

Solution

1. Step 1: Analyze traversal order in code

   The function calls postorder on right child first, then left child, then prints root.

2. Step 2: Compare with correct postorder order

   Correct postorder is left, right, root. This code reverses left and right order.

3. Final Answer:

   It visits right subtree before left, reversing postorder -> Option C

4. Quick Check:

   Left before right in postorder [OK]

Hint: Postorder = left first, then right, then root [OK]

Common Mistakes:

• Swapping left and right subtree calls
• Printing before visiting children
• Ignoring null checks

5. You want to collect all node values of a binary tree in postorder traversal into an array using TypeScript. Which code snippet correctly implements this?

hard

A. function postorder(node: TreeNode | null, result: number[]): void {
if (!node) return;
result.push(node.val);
postorder(node.left, result);
postorder(node.right, result);
}

B. function postorder(node: TreeNode | null, result: number[]): void {
if (!node) return;
postorder(node.left, result);
postorder(node.right, result);
result.push(node.val);
}

C. function postorder(node: TreeNode | null, result: number[]): number[] {
if (!node) return result;
result.push(node.val);
postorder(node.left, result);
postorder(node.right, result);
return result;
}

D. function postorder(node: TreeNode | null): number[] {
if (!node) return [];
return [node.val, ...postorder(node.left), ...postorder(node.right)];
}

Solution

1. Step 1: Understand postorder traversal collection

   Postorder visits left, then right, then root. Values should be pushed after children.

2. Step 2: Check each option for correct order and pushing

   function postorder(node: TreeNode | null, result: number[]): void { if (!node) return; postorder(node.left, result); postorder(node.right, result); result.push(node.val); } visits left, right, then pushes node value, correctly collecting postorder.

3. Final Answer:

   function postorder(node: TreeNode | null, result: number[]): void { if (!node) return; postorder(node.left, result); postorder(node.right, result); result.push(node.val); } -> Option B

4. Quick Check:

   Push after children = postorder [OK]

Hint: Push node value after left and right calls [OK]

Common Mistakes:

• Pushing node value before children
• Returning new arrays instead of using passed array
• Swapping left and right subtree calls