Tree Traversal Postorder Left Right Root in DSA Python - 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 the tree has more nodes?

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

This code visits every node in a tree in postorder: left child, right child, then the node itself.

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 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 longer because it visits nodes multiple times."

[OK] Correct: Each node is visited only once, just in a different order, so time grows linearly.

Interview Connect

Understanding traversal time helps you explain how tree algorithms scale and shows you can analyze recursive code clearly.

Self-Check

"What if we changed the traversal to 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. Root node, Left subtree, Right subtree

B. Right subtree, Left subtree, Root node

C. Root node, Right subtree, Left subtree

D. Left subtree, Right subtree, Root node

Solution

1. Step 1: Understand postorder traversal definition

   Postorder traversal visits nodes in the order: left subtree first, then 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 D

4. Quick Check:

   Postorder = Left, Right, Root [OK]

Hint: Remember: postorder ends with the 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 Python function headers correctly defines a postorder traversal method for a binary tree node root?

easy

A. def postorder(root):

B. def preorder(root):

C. def inorder(root):

D. def traverse(root):

Solution

1. Step 1: Identify the function name for postorder traversal

   The function name should clearly indicate postorder traversal, so postorder is correct.

2. Step 2: Check the parameter

   The parameter root is standard for the current node in tree traversal functions.

3. Final Answer:

   def postorder(root): -> Option A

4. Quick Check:

   Function name matches traversal type [OK]

Hint: Function name should match traversal type exactly [OK]

Common Mistakes:

• Using wrong traversal names like preorder or inorder
• Missing the root parameter
• Using generic names that don't specify traversal

3. Given the binary tree:

    1
   / \
  2   3
 / \
4   5

What is the output of the postorder traversal?

medium

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

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

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 of root (1)

   Right subtree rooted at 3 has no children, so postorder is [3]

3. Step 3: Combine left, right, and root

   Full postorder: left subtree [4, 5, 2], right subtree [3], root [1] -> [4, 5, 2, 3, 1]

4. Final Answer:

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

5. Quick Check:

   Postorder = Left, Right, Root [OK]

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

Common Mistakes:

• Visiting root before children
• Mixing up left and right subtree order
• Listing nodes in preorder or inorder instead

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

def postorder(root):
    if root:
        print(root.data)
        postorder(root.left)
        postorder(root.right)

medium

A. The left and right subtree calls are swapped

B. The function does not check if root is None

C. The root node is printed before traversing children

D. The function should return a list instead of printing

Solution

1. Step 1: Analyze the order of print and recursive calls

   The code prints root.data first, then traverses left and right subtrees. This is preorder traversal order.

2. Step 2: Identify correct postorder sequence

   Postorder requires visiting left subtree, then right subtree, then printing root last.

3. Final Answer:

   The root node is printed before traversing children -> Option C

4. Quick Check:

   Print after children calls for postorder [OK]

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

Common Mistakes:

• Printing root before children (preorder mistake)
• Not handling empty nodes properly
• Swapping left and right subtree calls

5. You want to collect all node values of a binary tree in postorder traversal into a list. Which code snippet correctly implements this?

hard

A. def postorder(root):
result = []
if root:
result.append(root.data)
result += postorder(root.left)
result += postorder(root.right)
return result

B. def postorder(root):
result = []
if root:
result += postorder(root.left)
result += postorder(root.right)
result.append(root.data)
return result

C. def postorder(root):
if root is None:
return []
return [root.data] + postorder(root.left) + postorder(root.right)

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

Solution

1. Step 1: Check the order of appending nodes

   def postorder(root): result = [] if root: result += postorder(root.left) result += postorder(root.right) result.append(root.data) return result appends left subtree results, then right subtree results, then root data last, matching postorder.

2. Step 2: Verify return of combined list

   def postorder(root): result = [] if root: result += postorder(root.left) result += postorder(root.right) result.append(root.data) return result returns the combined list correctly after recursive calls.

3. Final Answer:

   def postorder(root): result = [] if root: result += postorder(root.left) result += postorder(root.right) result.append(root.data) return result -> Option B

4. Quick Check:

   Left + Right + Root order in list [OK]

Hint: Append root after left and right recursive results [OK]

Common Mistakes:

• Appending root before children (preorder)
• Incorrect list concatenation order
• Not returning the combined list properly