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?"