Tree Traversal Preorder Root Left Right in DSA Javascript - Time & Space Complexity

Time Complexity: Tree Traversal Preorder Root Left Right

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 the tree has more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function preorderTraversal(node) {
  if (node === null) return;
  console.log(node.value); // Visit root
  preorderTraversal(node.left); // Traverse left subtree
  preorderTraversal(node.right); // Traverse right subtree
}
    

This code visits each node in the tree starting from the root, then left child, then right child.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Visiting each node once (printing its value).
• How many times: Exactly once per node in the tree.

How Execution Grows With Input

As the number of nodes increases, the steps increase directly in proportion.

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

Pattern observation: The number of steps grows linearly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

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

Common Mistake

[X] Wrong: "Preorder traversal takes more time because it visits the root first and then children."

[OK] Correct: The order of visiting nodes does not add extra steps; each node is still visited once, so time depends only on total nodes.

Interview Connect

Understanding preorder traversal time helps you explain how tree algorithms scale, a key skill in many coding challenges.

Self-Check

What if we changed preorder traversal to inorder traversal? How would the time complexity change?

Practice

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

easy

A. Left, Right, Root

B. Left, Root, Right

C. Root, Left, Right

D. Right, Root, Left

Solution

1. Step 1: Understand preorder traversal definition

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

2. Step 2: Confirm the visiting order

   The visiting order is Root, then Left child, then Right child.

3. Final Answer:

   Root, Left, Right -> Option C

4. Quick Check:

   Preorder = Root, Left, Right [OK]

Hint: Remember preorder starts with the root node first [OK]

Common Mistakes:

• Confusing preorder with inorder or postorder
• Visiting left or right child before root
• Mixing up left and right subtree order

2. Which of the following is the correct JavaScript function signature for preorder traversal of a binary tree node named root?

easy

A. function preorder(root) { if(root) { preorder(root.left); preorder(root.right); console.log(root.value); } }

B. function preorder(root) { if(root) { preorder(root.left); console.log(root.value); preorder(root.right); } }

C. function preorder(root) { if(root) { preorder(root.right); preorder(root.left); console.log(root.value); } }

D. function preorder(root) { if(root) { console.log(root.value); preorder(root.left); preorder(root.right); } }

Solution

1. Step 1: Recall preorder traversal order

   Preorder visits root first, then left subtree, then right subtree.

2. Step 2: Check function steps

   function preorder(root) { if(root) { console.log(root.value); preorder(root.left); preorder(root.right); } } prints root value first, then recursively visits left and right children, matching preorder.

3. Final Answer:

   function preorder(root) { if(root) { console.log(root.value); preorder(root.left); preorder(root.right); } } -> Option D

4. Quick Check:

   Print root before children = function preorder(root) { if(root) { console.log(root.value); preorder(root.left); preorder(root.right); } } [OK]

Hint: Print root value before recursive calls for preorder [OK]

Common Mistakes:

• Printing after visiting children (postorder)
• Swapping left and right recursive calls
• Missing base case for null root

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

    1
   / \
  2   3
 / \
4   5

Use the format: 1 -> 2 -> 4 -> 5 -> 3 -> null

medium

A. 1 -> 2 -> 4 -> 5 -> 3 -> null

B. 4 -> 2 -> 5 -> 1 -> 3 -> null

C. 2 -> 4 -> 5 -> 1 -> 3 -> null

D. 1 -> 3 -> 2 -> 4 -> 5 -> null

Solution

1. Step 1: Start at root node 1

   Visit root first: 1

2. Step 2: Traverse left subtree of 1

   Visit 2, then its left child 4, then right child 5

3. Step 3: Traverse right subtree of 1

   Visit 3

4. Final Answer:

   1 -> 2 -> 4 -> 5 -> 3 -> null -> Option A

5. Quick Check:

   Preorder visits root, left, right = 1 -> 2 -> 4 -> 5 -> 3 -> null [OK]

Hint: Follow root, then left subtree, then right subtree order [OK]

Common Mistakes:

• Visiting left subtree after right subtree
• Starting traversal from left child instead of root
• Confusing preorder with inorder or postorder

4. The following JavaScript code is intended to perform a preorder traversal and print node values. What is the error?

function preorder(root) {
  if (root === null) return;
  preorder(root.left);
  console.log(root.value);
  preorder(root.right);
}

medium

A. The console.log should be before the recursive calls

B. The base case should check for undefined, not null

C. The function should return a value instead of void

D. The left and right recursive calls are swapped

Solution

1. Step 1: Analyze the order of operations

   The code visits left child first, then prints root, then visits right child.

2. Step 2: Compare with preorder traversal order

   Preorder requires printing root before visiting children, so console.log should be before recursive calls.

3. Final Answer:

   The console.log should be before the recursive calls -> Option A

4. Quick Check:

   Print root first in preorder [OK]

Hint: Print root value before recursive calls in preorder [OK]

Common Mistakes:

• Printing after left subtree instead of before
• Confusing preorder with inorder traversal
• Ignoring base case for null nodes

5. You have a binary tree where some nodes have only one child. How does preorder traversal handle nodes with a missing left child?
Consider this tree:

    10
      \
      20
     /  \
   30    40

What is the preorder traversal output?

hard

A. 10 -> 40 -> 20 -> 30 -> null

B. 10 -> 20 -> 30 -> 40 -> null

C. 10 -> 20 -> 40 -> 30 -> null

D. 10 -> 30 -> 20 -> 40 -> null

Solution

1. Step 1: Visit root node 10

   Start with root: 10

2. Step 2: Check left child of 10

   Left child is missing, so skip to right child 20

3. Step 3: Visit node 20 and its children

   Visit 20, then left child 30, then right child 40

4. Final Answer:

   10 -> 20 -> 30 -> 40 -> null -> Option B

5. Quick Check:

   Preorder visits root, left (if any), right = 10 -> 20 -> 30 -> 40 -> null [OK]

Hint: Skip missing children, visit existing nodes in preorder order [OK]

Common Mistakes:

• Assuming missing left child means skip right child
• Visiting children in wrong order
• Confusing preorder with other traversals