Vertical Order Traversal of Binary Tree in DSA Javascript - 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?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function verticalOrder(root) {
  if (!root) return [];
  const map = new Map();
  const queue = [[root, 0]];
  while (queue.length) {
    const [node, col] = queue.shift();
    if (!map.has(col)) map.set(col, []);
    map.get(col).push(node.val);
    if (node.left) queue.push([node.left, col - 1]);
    if (node.right) queue.push([node.right, col + 1]);
  }
  return [...map.entries()].sort((a, b) => a[0] - b[0]).map(entry => entry[1]);
}
    

This code does a breadth-first search to group nodes by their vertical column index.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop processes each node once.
• How many times: Exactly once per node, so n times if there are n nodes.

How Execution Grows With Input

As the number of nodes grows, the loop runs once for each node.

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

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

Final Time Complexity

Time Complexity: O(n + k log k)

This means the time grows a bit faster than the number of nodes because sorting the columns takes extra time.

Common Mistake

[X] Wrong: "The traversal is just O(n) because we visit each node once."

[OK] Correct: We also sort the column keys at the end, which takes O(k log k) time, where k is the number of columns, making total time O(n + k log k). Since k can be up to n, this is O(n log n).

Interview Connect

Understanding this time complexity helps you explain how traversal and sorting steps combine, showing you can analyze multi-step algorithms clearly.

Self-Check

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

Practice

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

easy

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

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

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

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

Solution

1. Step 1: Understand vertical columns in a binary tree

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

2. Step 2: Identify traversal order

   Nodes in the same vertical column are collected and output from leftmost to rightmost column.

3. Final Answer:

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

4. Quick Check:

   Vertical order = group by columns [OK]

Hint: Think columns, not rows or preorder traversal [OK]

Common Mistakes:

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

2. Which of the following JavaScript code snippets correctly initializes a map to store vertical order nodes by column?

easy

A. const map = new Map();

B. const map = {};

C. const map = new Set();

D. const map = [];

Solution

1. Step 1: Identify suitable data structure for key-value pairs

   We need to map column numbers (keys) to arrays of nodes (values), so Map is ideal.

2. Step 2: Check each option

   const map = new Map(); uses Map which supports keys of any type and preserves insertion order. const map = {}; is an object but less flexible for numeric keys. const map = new Set(); is a Set, which stores unique values without keys. const map = []; is an array, not suitable for key-value mapping.

3. Final Answer:

   const map = new Map(); -> Option A

4. Quick Check:

   Use Map for column to nodes mapping [OK]

Hint: Use Map for flexible key-value storage [OK]

Common Mistakes:

• Using Set instead of Map for key-value pairs
• Using array instead of Map for mapping columns
• Using plain object which may cause key type issues

3. Given the binary tree:

    1
   / \
  2   3
 / \   \
4   5   6

What is the vertical order traversal output?

medium

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

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

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

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

Solution

1. Step 1: Assign column indices to nodes

   Root (1) is column 0. Left child decreases column by 1, right child increases by 1.
   So, 4 is at -2, 2 at -1, 1 and 5 at 0, 3 at 1, 6 at 2.

2. Step 2: Group nodes by columns and list them 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 A

4. Quick Check:

   Column grouping matches answer A [OK]

Hint: Track column index: left -1, right +1 [OK]

Common Mistakes:

• Mixing node order within the same column
• Ignoring column indices and grouping by levels
• Swapping left and right child column adjustments

4. Consider this code snippet for vertical order traversal:

function verticalOrder(root) {
  const map = new Map();
  function dfs(node, col) {
    if (!node) return;
    dfs(node.left, col - 1);
    dfs(node.right, col + 1);
    if (!map.has(col)) map.set(col, []);
    map.get(col).push(node.val);
  }
  dfs(root, 0);
  return [...map.values()];
}

What is the main issue with this code?

medium

A. The function does not handle empty trees.

B. Map keys should be strings, not numbers.

C. DFS does not guarantee top-to-bottom order in vertical columns.

D. The code uses recursion incorrectly.

Solution

1. Step 1: Understand DFS traversal order

   DFS explores depth first, so nodes deeper in the tree may be visited before nodes higher up in the same column.

2. Step 2: Identify vertical order requirement

   Vertical order requires nodes in each column to be output top to bottom, which DFS alone does not guarantee.

3. Final Answer:

   DFS does not guarantee top-to-bottom order in vertical columns. -> Option C

4. Quick Check:

   DFS order ≠ vertical top-to-bottom order [OK]

Hint: Use BFS for correct vertical top-to-bottom order [OK]

Common Mistakes:

• Assuming DFS preserves vertical order top-to-bottom
• Thinking Map keys must be strings
• Ignoring empty tree base case (which is handled here)

5. You want to perform vertical order traversal of a binary tree where nodes in the same column and row are sorted by value ascending. Which approach correctly achieves this?

hard

A. Use DFS and store nodes in a map by column only, then output map values.

B. Use BFS with a queue storing nodes with their row and column, then sort nodes by column, row, and value.

C. Traverse tree inorder and group nodes by column index only.

D. Use BFS but ignore row information and output nodes by column only.

Solution

1. Step 1: Understand sorting requirements

   Nodes must be sorted by column, then row (top to bottom), then by node value ascending if same row and column.

2. Step 2: Choose traversal and data structure

   BFS with queue tracks nodes level by level (row), and storing column and row allows sorting by these keys plus value.

3. Step 3: Why other options fail

   DFS does not guarantee row order. Inorder traversal does not track columns properly. Ignoring row info loses sorting order.

4. Final Answer:

   Use BFS with a queue storing nodes with their row and column, then sort nodes by column, row, and value. -> Option B

5. Quick Check:

   BFS + sort by col, row, value = correct order [OK]

Hint: Track row and column with BFS, then sort by col, row, value [OK]

Common Mistakes:

• Using DFS which loses row order
• Not sorting nodes by value when row and column are same
• Ignoring row info in BFS