Tree Traversal Preorder Root Left Right in DSA Typescript - 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(root: TreeNode | null): number[] {
  if (root === null) return [];
  const left = preorderTraversal(root.left);
  const right = preorderTraversal(root.right);
  return [root.val, ...left, ...right];
}
    

This code visits each node in a tree starting from the root, then left child, then right child, collecting values in that order.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

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

How Execution Grows With Input

Each node is visited exactly once, so the 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 time grows linearly as the tree size increases.

Final Time Complexity

Time Complexity: O(n)

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

Common Mistake

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

[OK] Correct: Each node is visited exactly once, so the total steps grow only with the number of nodes, not more.

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. Root, Left, Right

B. Left, Root, Right

C. Left, Right, Root

D. Right, Left, Root

Solution

1. Step 1: Understand preorder traversal definition

   Preorder traversal visits the root node first, then recursively visits the left subtree, followed by the right subtree.

2. Step 2: Match the order with options

   The order Root, Left, Right matches the preorder traversal definition exactly.

3. Final Answer:

   Root, Left, Right -> Option A

4. Quick Check:

   Preorder = Root, Left, Right [OK]

Hint: Preorder always starts with the root node first [OK]

Common Mistakes:

• Confusing preorder with inorder or postorder
• Visiting left or right before root
• Mixing up traversal orders

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

easy

A. function preorder(node: TreeNode): number[]

B. function preorder(node: TreeNode | null): void

C. function preorder(): TreeNode

D. function preorder(node: TreeNode, result: number[]): number

Solution

1. Step 1: Analyze function parameters and return type

   A preorder traversal function typically takes a node (or null) and returns void if it prints or processes nodes directly.

2. Step 2: Check each option

   function preorder(node: TreeNode | null): void accepts a node or null and returns void, which fits a traversal that processes nodes without returning a value.

3. Final Answer:

   function preorder(node: TreeNode | null): void -> Option B

4. Quick Check:

   Traversal function signature with node or null and void return [OK]

Hint: Traversal functions usually take a node and return void or array [OK]

Common Mistakes:

• Missing null in node parameter type
• Wrong return type for traversal
• Incorrect number of parameters

3. Given the following binary tree:

    1
   / \
  2   3
 / \
4   5

What is the output of a preorder traversal (Root, Left, Right)?

medium

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

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

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

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

Solution

1. Step 1: Start at root node 1

   Preorder visits root first: 1

2. Step 2: Traverse left subtree of 1

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

3. Step 3: Traverse right subtree of 1

   Visit node 3: 3

4. Final Answer:

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

5. Quick Check:

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

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

Common Mistakes:

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

4. Consider this TypeScript preorder traversal function:

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

What is the error in this code?

medium

A. It does not handle null nodes

B. It will cause infinite recursion

C. It prints nodes in postorder

D. It visits nodes in inorder, not preorder

Solution

1. Step 1: Analyze the order of operations in the function

   The function first calls preorder on left child, then prints current node, then calls preorder on right child.

2. Step 2: Compare with preorder definition

   Preorder should print the node first, then traverse left and right. This code prints after left subtree, so it is inorder traversal.

3. Final Answer:

   It visits nodes in inorder, not preorder -> Option D

4. Quick Check:

   Print before recursive calls = preorder; here print is in middle [OK]

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

Common Mistakes:

• Printing node after left subtree instead of before
• Confusing traversal orders
• Missing base case for null

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

hard

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

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

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

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

Solution

1. Step 1: Understand preorder traversal collection

   Preorder visits root first, then left subtree, then right subtree, collecting values in order.

2. Step 2: Check each function's order and correctness

   function preorder(node: TreeNode | null, result: number[] = []): number[] { if (!node) return result; result.push(node.val); preorder(node.left, result); preorder(node.right, result); return result; } pushes node.val first, then recursively collects left and right, returning the result array correctly.

3. Final Answer:

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

4. Quick Check:

   Push root first, then left and right recursively [OK]

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

Common Mistakes:

• Collecting left subtree before root
• Mixing left and right subtree order
• Returning new arrays instead of accumulating