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

Time Complexity: Tree Traversal Preorder Root Left Right

O(n)

Understanding Time Complexity

We want to understand how long it takes to visit every node in a tree using preorder traversal.

How does the time grow as the tree gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

def preorder(root):
    if root is None:
        return
    print(root.val)
    preorder(root.left)
    preorder(root.right)

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 and printing its value.
• How many times: Exactly once per node in the tree.

How Execution Grows With Input

As the number of nodes grows, the number of visits grows the same way.

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

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

Final Time Complexity

Time Complexity: O(n)

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

Common Mistake

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

[OK] Correct: Each node is visited exactly once, so the time grows only with the number of nodes.

Interview Connect

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

Self-Check

"What if we changed preorder 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. Right, Root, Left

B. Left, Root, Right

C. Root, Left, Right

D. Left, Right, Root

Solution

1. Step 1: Understand preorder traversal definition

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

2. Step 2: Match the order with options

   Root, Left, Right matches the order Root, Left, Right exactly as defined.

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 subtree before root
• Mixing up left and right subtree order

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

easy

A. def preorder(root):

B. def preorder_traversal():

C. def preorder(root, left, right):

D. def preorder(root, node):

Solution

1. Step 1: Identify function parameters for preorder traversal

   The function needs the root node to start traversal, so it should accept one parameter named root.

2. Step 2: Check options for correct syntax

   def preorder(root): defines a function with one parameter root, which is correct and simple for preorder traversal.

3. Final Answer:

   def preorder(root): -> Option A

4. Quick Check:

   Function header with root parameter [OK]

Hint: Preorder function usually takes only the root node [OK]

Common Mistakes:

• Adding unnecessary parameters
• Missing the root parameter
• Incorrect function syntax

3. Given the binary tree:

    1
   / \
  2   3
 / \
4   5

What is the output of the preorder traversal?

medium

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

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

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

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

Solution

1. Step 1: Start preorder traversal at root node 1

   Visit 1 first, then move to left subtree.

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 last.

4. Final Answer:

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

5. Quick Check:

   Preorder visits 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 before root
• Swapping left and right children
• Confusing preorder with inorder or postorder

4. Consider this preorder traversal function:

def preorder(root):
    if root is None:
        return
    print(root.val)
    preorder(root.right)
    preorder(root.left)

What is the error in this code?

medium

A. It misses printing the root node

B. It does not check if root is None

C. It prints node values twice

D. It visits right subtree before left subtree, reversing preorder order

Solution

1. Step 1: Analyze the order of recursive calls

   The function calls preorder on root.right before root.left, which reverses the expected left-right order.

2. Step 2: Understand preorder traversal order

   Preorder must visit root, then left subtree, then right subtree. This code visits right subtree before left subtree, which is incorrect.

3. Final Answer:

   It visits right subtree before left subtree, reversing preorder order -> Option D

4. Quick Check:

   Left subtree must come before right in preorder [OK]

Hint: Left subtree must be visited before right in preorder [OK]

Common Mistakes:

• Swapping left and right recursive calls
• Forgetting base case check
• Printing nodes multiple times

5. You want to collect all node values in preorder traversal into a list. Which code snippet correctly does this for a binary tree node root?

hard

A. def preorder(root):
if root is None:
return []
return [root.val] + preorder(root.left) + preorder(root.right)

B. def preorder(root):
result = []
if root:
preorder(root.left)
result.append(root.val)
preorder(root.right)
return result

C. def preorder(root):
result = []
if root:
result.append(root.val)
result.extend(preorder(root.left))
result.extend(preorder(root.right))
return result

D. def preorder(root):
if root is None:
return []
return preorder(root.left) + [root.val] + preorder(root.right)

Solution

1. Step 1: Understand preorder traversal collection

   Preorder visits root first, then left subtree, then right subtree. We must collect values in this order.

2. Step 2: Analyze each option's return order

   def preorder(root): if root is None: return [] return [root.val] + preorder(root.left) + preorder(root.right) returns [root.val] + preorder(left) + preorder(right), which matches preorder order correctly.

3. Step 3: Identify errors in other options

   def preorder(root): result = [] if root: result.append(root.val) result.extend(preorder(root.left)) result.extend(preorder(root.right)) return result uses result list but calls preorder recursively without passing result, causing empty lists. def preorder(root): result = [] if root: preorder(root.left) result.append(root.val) preorder(root.right) return result appends root after left subtree, which is inorder. def preorder(root): if root is None: return [] return preorder(root.left) + [root.val] + preorder(root.right) places root.val between left and right, which is inorder.

4. Final Answer:

   def preorder(root): if root is None: return [] return [root.val] + preorder(root.left) + preorder(root.right) -> Option A

5. Quick Check:

   Preorder = root + left + right [OK]

Hint: Return root value first, then left and right lists concatenated [OK]

Common Mistakes:

• Appending root after left subtree (inorder)
• Not returning combined list properly
• Using mutable list without passing it recursively