Tree Traversal Postorder Left Right Root in DSA Javascript - 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.

How does the number of steps change when we use postorder traversal?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function postorderTraversal(node) {
  if (!node) return;
  postorderTraversal(node.left);
  postorderTraversal(node.right);
  console.log(node.value);
}
    

This code visits every node in a tree by first going left, then right, and finally processing the current node.

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

Each node is visited once, so the total steps grow 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 finish grows directly with the number of nodes in the tree.

Common Mistake

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

[OK] Correct: Each node is visited exactly once, so the time grows linearly, not more.

Interview Connect

Understanding this helps you explain how tree traversals work efficiently and shows you can analyze recursive code clearly.

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 order with options

   Left subtree, Right subtree, Root node matches the order Left, Right, Root exactly.

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 node [OK]

Common Mistakes:

• Confusing postorder with preorder or inorder
• Visiting root node first instead of last
• Swapping left and right subtree order

2. Which of the following is the correct syntax to perform a postorder traversal function call in JavaScript assuming postorder(node) is defined?

easy

A. console.log(node.value); postorder(node.left); postorder(node.right);

B. console.log(node.value); postorder(node.right); postorder(node.left);

C. postorder(node.right); postorder(node.left); console.log(node.value);

D. postorder(node.left); postorder(node.right); console.log(node.value);

Solution

1. Step 1: Recall postorder traversal steps

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

2. Step 2: Check syntax order

   postorder(node.left); postorder(node.right); console.log(node.value); calls postorder on left, then right, then prints root value last, matching postorder.

3. Final Answer:

   postorder(node.left); postorder(node.right); console.log(node.value); -> Option D

4. Quick Check:

   Left, Right, Root call order = postorder(node.left); postorder(node.right); console.log(node.value); [OK]

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

Common Mistakes:

• Printing root before recursive calls
• Swapping left and right recursive calls
• Missing recursive calls for subtrees

3. Given the binary tree below, what is the output of a postorder traversal?

    1
   / \
  2   3
 / \
4   5

medium

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

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

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 is left(4), right(5), root(2) -> [4, 5, 2]

2. Step 2: Traverse right subtree and root

   Right subtree is node 3 -> [3]. Then root 1 last -> [1]. Combine: [4, 5, 2] + [3] + [1] = [4, 5, 2, 3, 1]

3. Final Answer:

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

4. Quick Check:

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

Hint: Visit left subtree fully before right, then root [OK]

Common Mistakes:

• Listing root before subtrees
• Mixing up left and right subtree order
• Confusing preorder or inorder output

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

function postorder(node) {
  if (!node) return;
  console.log(node.value);
  postorder(node.left);
  postorder(node.right);
}

medium

A. The left subtree is visited after the right subtree

B. The function does not check if node is null

C. The root node is printed before its children, not after

D. The function is missing a return statement

Solution

1. Step 1: Analyze print order in code

   The code prints node.value before recursive calls, so root prints first.

2. Step 2: Compare with postorder definition

   Postorder requires printing root after left and right subtrees, so printing first is wrong.

3. Final Answer:

   The root node is printed before its children, not after -> Option C

4. Quick Check:

   Postorder prints root last, code prints root first [OK]

Hint: Print root after recursive calls for postorder [OK]

Common Mistakes:

• Printing root before recursive calls
• Ignoring base case for null nodes
• Swapping left and right subtree calls

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

hard

A. function postorder(node, arr = []) {
if (!node) return arr;
postorder(node.left, arr);
postorder(node.right, arr);
arr.push(node.value);
return arr;
}

B. function postorder(node, arr = []) {
if (!node) return arr;
arr.push(node.value);
postorder(node.left, arr);
postorder(node.right, arr);
return arr;
}

C. function postorder(node, arr = []) {
if (!node) return arr;
postorder(node.right, arr);
postorder(node.left, arr);
arr.push(node.value);
return arr;
}

D. function postorder(node, arr = []) {
if (!node) return arr;
postorder(node.left, arr);
arr.push(node.value);
postorder(node.right, arr);
return arr;
}

Solution

1. Step 1: Check traversal order in each snippet

   function postorder(node, arr = []) { if (!node) return arr; postorder(node.left, arr); postorder(node.right, arr); arr.push(node.value); return arr; } visits left subtree, then right subtree, then pushes root value last, matching postorder.

2. Step 2: Verify array accumulation and return

   function postorder(node, arr = []) { if (!node) return arr; postorder(node.left, arr); postorder(node.right, arr); arr.push(node.value); return arr; } correctly passes and returns the array, accumulating values in postorder.

3. Final Answer:

   Option A code snippet -> Option A

4. Quick Check:

   Left, Right, Root with arr.push last = function postorder(node, arr = []) { if (!node) return arr; postorder(node.left, arr); postorder(node.right, arr); arr.push(node.value); return arr; } [OK]

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

Common Mistakes:

• Pushing root value before recursive calls
• Swapping left and right subtree calls
• Not returning the array properly