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

Time Complexity: Vertical Order Traversal of Binary Tree

O(n log k)

Understanding Time Complexity

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

Specifically, how does the number of nodes affect the work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public IList> VerticalOrder(TreeNode root) {
    var result = new List>();
    if (root == null) return result;
    var map = new SortedDictionary>();
    var queue = new Queue<(TreeNode node, int col)>();
    queue.Enqueue((root, 0));
    while (queue.Count > 0) {
        var (node, col) = queue.Dequeue();
        if (!map.ContainsKey(col)) map[col] = new List();
        map[col].Add(node.val);
        if (node.left != null) queue.Enqueue((node.left, col - 1));
        if (node.right != null) queue.Enqueue((node.right, col + 1));
    }
    foreach (var key in map.Keys) result.Add(map[key]);
    return result;
}
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Visiting each node once in a breadth-first manner.
• How many times: Exactly once per node, so n times for n nodes.
• Also, inserting and accessing the map for each node's column key.
• Finally, iterating over the sorted keys to build the result.

How Execution Grows With Input

As the number of nodes grows, the code visits each node once and does a few constant-time operations per node.

Input Size (n)	Approx. Operations
10	About 10 visits plus map insertions and lookups
100	About 100 visits plus map operations
1000	About 1000 visits plus map operations

Pattern observation: The work grows roughly in direct proportion to the number of nodes.

Final Time Complexity

Time Complexity: O(n log k)

This means the time grows mostly with the number of nodes, but also depends on the number of vertical columns (k) because of sorting keys.

Common Mistake

[X] Wrong: "The traversal is just O(n) because we visit each node once, so sorting keys doesn't matter."

[OK] Correct: The keys for columns are stored in a sorted structure, so accessing and inserting keys costs extra time depending on how many columns there are.

Interview Connect

Understanding how tree traversal and data grouping affect time helps you explain your approach clearly and shows you know how to analyze your code's efficiency.

Self-Check

"What if we used a hash map instead of a sorted dictionary for columns? How would the time complexity change?"

Practice

1. What does the vertical order traversal of a binary tree represent?

easy

A. Nodes grouped by their depth level from top to bottom

B. Nodes grouped by their vertical columns from left to right

C. Nodes visited in preorder (root, left, right)

D. Nodes visited in postorder (left, right, root)

Solution

1. Step 1: Understand vertical order traversal

   Vertical order traversal groups nodes by their vertical columns, where the root is at column 0, left child at column -1, right child at column +1, and so on.

2. Step 2: Compare with other traversals

   Unlike level order or preorder, vertical order focuses on columns, not depth or visit order.

3. Final Answer:

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

4. Quick Check:

   Vertical order = nodes by vertical columns [OK]

Hint: Vertical order groups nodes by columns, not levels or visit order [OK]

Common Mistakes:

• Confusing vertical order with level order traversal
• Thinking vertical order is preorder or postorder
• Ignoring the horizontal distance concept

2. Which of the following C# code snippets correctly initializes a dictionary to store vertical order traversal results where keys are integers and values are lists of integers?

easy

A. var dict = new Dictionary();

B. var dict = new Dictionary();

C. var dict = new Dictionary();

D. var dict = new List>();

Solution

1. Step 1: Identify the data structure needed

   We need a dictionary with integer keys (vertical columns) and list of integers as values (nodes in that column).

2. Step 2: Match with C# syntax

   var dict = new Dictionary(); correctly declares Dictionary. Others have wrong types or collections.

3. Final Answer:

   var dict = new Dictionary(); -> Option A

4. Quick Check:

   Dictionary for vertical order [OK]

Hint: Use Dictionary for vertical columns and nodes [OK]

Common Mistakes:

• Swapping key and value types
• Using List instead of Dictionary for mapping
• Incorrect generic type order

3. Given the following binary tree and C# code for vertical order traversal, what is the printed output?

    1
   / \
  2   3
 / \   \
4   5   6

// Code snippet:
var result = new Dictionary();
void Traverse(TreeNode node, int col) {
  if (node == null) return;
  if (!result.ContainsKey(col)) result[col] = new List<int>();
  result[col].Add(node.val);
  Traverse(node.left, col - 1);
  Traverse(node.right, col + 1);
}
Traverse(root, 0);
foreach (var col in result.OrderBy(k => k.Key)) {
  Console.Write(string.Join(",", col.Value) + " ");
}

medium

A. 4 2 1 5 3 6

B. 4,2 1,5,3 6

C. 4 2 1,5 3 6

D. 4,2 1 5,3 6

Solution

1. Step 1: Assign columns to nodes

   Root 1 at col 0; 2 at -1; 3 at +1; 4 at -2; 5 at 0; 6 at +2.

2. Step 2: Group nodes by column and order

   col -2: [4], col -1: [2], col 0: [1,5], col 1: [3], col 2: [6]. Printed as 4 2 1,5 3 6 (commas inside multi-node groups, spaces between columns).

3. Final Answer:

   4 2 1,5 3 6 -> Option C

4. Quick Check:

   Vertical groups with commas inside, spaces between [OK]

Hint: Track column indices and group nodes before printing [OK]

Common Mistakes:

• Mixing node order inside columns
• Printing without commas or spaces correctly
• Ignoring column sorting

4. The following C# code attempts vertical order traversal but throws a runtime error. What is the main issue?

void Traverse(TreeNode node, int col, Dictionary dict) {
  if (node == null) return;
  dict[col].Add(node.val);
  Traverse(node.left, col - 1, dict);
  Traverse(node.right, col + 1, dict);
}

var dict = new Dictionary();
Traverse(root, 0, dict);

medium

A. Dictionary should use List<int> as key

B. Traverse method should return a value

C. Incorrect base case for recursion

D. Dictionary keys are not initialized before adding values

Solution

1. Step 1: Analyze dictionary usage

   Code tries to add to dict[col] without checking if key col exists, causing KeyNotFoundException.

2. Step 2: Fix by initializing keys

   Before adding, check if dict contains col; if not, create new List<int> for that key.

3. Final Answer:

   Dictionary keys are not initialized before adding values -> Option D

4. Quick Check:

   Initialize dictionary keys before adding [OK]

Hint: Always check and initialize dictionary keys before adding [OK]

Common Mistakes:

• Assuming dictionary auto-creates keys
• Ignoring null checks on dictionary values
• Confusing recursion base case with dictionary errors

5. You want to perform vertical order traversal but also sort nodes by their depth (top to bottom) within each vertical column. Which data structure and approach is best in C#?

hard

A. Use a Dictionary> and sort each list by depth after traversal

B. Use a Dictionary and enqueue nodes as visited

C. Use a List<int> and sort by node value after traversal

D. Use a Dictionary and push nodes in preorder

Solution

1. Step 1: Understand sorting by depth requirement

   We must store nodes with their depth info to sort them top to bottom within each vertical column.

2. Step 2: Choose data structure and approach

   Dictionary with key as column and value as list of (depth, value) pairs allows sorting by depth after traversal.

3. Final Answer:

   Use a Dictionary> and sort each list by depth after traversal -> Option A

4. Quick Check:

   Store depth with nodes, then sort by depth [OK]

Hint: Store depth with nodes, then sort lists by depth per column [OK]

Common Mistakes:

• Using queue or stack without depth info
• Sorting by node value instead of depth
• Ignoring depth when grouping nodes