Vertical Order Traversal of Binary Tree in DSA Python - Time & Space Complexity

Time Complexity: Vertical Order Traversal of Binary Tree

O(n log n)

Understanding Time Complexity

We want to understand how the time needed to do vertical order traversal grows as the tree gets bigger.

Specifically, how does the number of nodes affect the work done to group nodes by their vertical columns?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

from collections import defaultdict, deque

def vertical_order_traversal(root):
    if not root:
        return []
    column_table = defaultdict(list)
    queue = deque([(root, 0)])
    while queue:
        node, column = queue.popleft()
        column_table[column].append(node.val)
        if node.left:
            queue.append((node.left, column - 1))
        if node.right:
            queue.append((node.right, column + 1))
    return [column_table[x] for x in sorted(column_table)]
    

This code visits each node once, grouping node values by their vertical column index.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that visits each node once.
• How many times: Exactly once per node, so n times for n nodes.
• Sorting the keys of the column table at the end, which depends on the number of columns.

How Execution Grows With Input

As the number of nodes grows, the traversal visits each node once, so work grows roughly in direct proportion to n.

Input Size (n)	Approx. Operations
10	About 10 visits plus sorting few columns
100	About 100 visits plus sorting columns (up to 100)
1000	About 1000 visits plus sorting columns (up to 1000)

Pattern observation: The main work grows linearly with nodes, and sorting columns grows with the number of distinct vertical columns, which is at most n.

Final Time Complexity

Time Complexity: O(n log n)

This means the time grows a bit faster than the number of nodes because of sorting the column keys, but the main traversal is linear.

Common Mistake

[X] Wrong: "The traversal alone takes O(n log n) because of the queue operations."

[OK] Correct: Each node is added and removed from the queue exactly once, so queue operations are O(n) total, not O(n log n). The sorting step causes the log n factor.

Interview Connect

Understanding this complexity helps you explain how tree traversals scale and how sorting affects overall time, a useful skill for many tree-related problems.

Self-Check

"What if we used a balanced tree or heap to keep columns sorted during traversal? How would the time complexity change?"

Practice

1. What does Vertical Order Traversal of a binary tree mean?

easy

A. Nodes are grouped by their horizontal rows from top to bottom

B. Nodes are visited level by level from top to bottom

C. Nodes are grouped by their vertical columns from left to right

D. Nodes are visited in preorder (root, left, right)

Solution

1. Step 1: Understand vertical columns in a tree

   Vertical order groups nodes by their horizontal distance from the root, called columns.

2. Step 2: Define vertical order traversal

   It means collecting nodes column-wise from leftmost to rightmost column.

3. Final Answer:

   Nodes are grouped by their vertical columns from left to right -> Option C

4. Quick Check:

   Vertical order = group nodes by columns [OK]

Hint: Think columns from left to right for vertical order [OK]

Common Mistakes:

• Confusing vertical order with level order traversal
• Thinking vertical order is preorder or inorder traversal
• Mixing rows with columns in traversal

2. Which Python data structure is best to store nodes grouped by their vertical columns during vertical order traversal?

easy

A. Dictionary with column as key and list of nodes as value

B. List

C. Set

D. Tuple

Solution

1. Step 1: Identify need to group nodes by column

   We need to map each column number to the nodes in that column.

2. Step 2: Choose data structure for mapping

   A dictionary with column keys and list values is ideal for grouping nodes by column.

3. Final Answer:

   Dictionary with column as key and list of nodes as value -> Option A

4. Quick Check:

   Use dict for column to nodes mapping [OK]

Hint: Use dict to map columns to node lists [OK]

Common Mistakes:

• Using list only without mapping columns
• Using set which loses order
• Using tuple which is immutable

3. What is the output of vertical order traversal for this tree?

    1
   / \
  2   3
 / \   \
4   5   6

medium

A. [[4, 2], [1, 5, 3], [6]]

B. [[4], [2], [1, 5], [3], [6]]

C. [[1], [2, 3], [4, 5, 6]]

D. [[6], [3], [1, 5], [2], [4]]

Solution

1. Step 1: Assign column indices to nodes

   Root 1 is column 0; left child 2 is -1; right child 3 is +1; 4 is -2; 5 is 0; 6 is +2.

2. Step 2: Group nodes by columns and order top to bottom

   Columns: -2:[4], -1:[2], 0:[1,5], 1:[3], 2:[6].

3. Final Answer:

   [[4], [2], [1, 5], [3], [6]] -> Option B

4. Quick Check:

   Group nodes by column indices [OK]

Hint: Assign columns: left -1, right +1 from root [OK]

Common Mistakes:

• Mixing node order within same column
• Ignoring column indices for children
• Grouping nodes by rows instead of columns

4. Identify the bug in this vertical order traversal code snippet:

def vertical_order(root):
    if not root:
        return []
    col_table = {}
    queue = [(root, 0)]
    while queue:
        node, col = queue.pop()
        if col not in col_table:
            col_table[col] = []
        col_table[col].append(node.val)
        if node.left:
            queue.append((node.left, col - 1))
        if node.right:
            queue.append((node.right, col + 1))
    return [col_table[x] for x in sorted(col_table)]

medium

A. Using pop() removes last element, should use pop(0) for BFS

B. Dictionary keys should be strings, not integers

C. Appending node values instead of nodes causes error

D. Missing check for empty tree at start

Solution

1. Step 1: Analyze queue usage for BFS

   Queue should remove elements from front (pop(0)) for BFS, but pop() removes from end (LIFO).

2. Step 2: Identify impact on traversal order

   Using pop() causes DFS-like order, breaking vertical order traversal logic.

3. Final Answer:

   Using pop() removes last element, should use pop(0) for BFS -> Option A

4. Quick Check:

   Use pop(0) for BFS queue [OK]

Hint: Use pop(0) to dequeue for BFS, not pop() [OK]

Common Mistakes:

• Using pop() instead of pop(0) in BFS queue
• Confusing BFS with DFS traversal order
• Not sorting dictionary keys before output

5. Given a binary tree where multiple nodes share the same vertical and horizontal distance, how should vertical order traversal output be ordered?

hard

A. Sort nodes by column only, ignoring row and value

B. Sort nodes by value only, ignoring position

C. Output nodes in insertion order without sorting

D. Sort nodes by their row (top to bottom), then by value ascending if tie

Solution

1. Step 1: Understand tie cases in vertical order

   When nodes share same column and row, order by node value ascending to maintain consistency.

2. Step 2: Define final ordering rule

   Sort nodes first by row (top to bottom), then by value ascending if rows tie.

3. Final Answer:

   Sort nodes by their row (top to bottom), then by value ascending if tie -> Option D

4. Quick Check:

   Row then value sorting for ties [OK]

Hint: Sort by row, then value for nodes sharing position [OK]

Common Mistakes:

• Ignoring row order and sorting only by column
• Not sorting nodes with same position by value
• Outputting nodes in random or insertion order