Vertical Order Traversal of Binary Tree in DSA Typescript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is Vertical Order Traversal of a Binary Tree?

It is a way to print the nodes of a binary tree column by column, from left to right. Nodes in the same vertical line are grouped together.

beginner

How do we assign vertical levels to nodes in Vertical Order Traversal?

The root node is at vertical level 0. Left child nodes decrease the vertical level by 1, right child nodes increase it by 1.

intermediate

Which data structure is commonly used to store nodes by their vertical levels during traversal?

A map or dictionary with keys as vertical levels and values as lists of nodes at that level.

intermediate

Why do we use a queue in Vertical Order Traversal?

A queue helps us do a level order traversal (breadth-first) so nodes are processed top to bottom within each vertical line.

advanced

What is the output order when multiple nodes share the same vertical and horizontal level?

They are printed in the order they appear in the level order traversal (top to bottom, left to right).

In Vertical Order Traversal, what vertical level is assigned to the root node?

A 1

B 0

C -1

D Depends on tree height

The root node is always assigned vertical level 0.

Which data structure is best to keep track of nodes and their vertical levels during traversal?

A Stack

B Queue

C Set

D Map/Dictionary

A map/dictionary stores vertical levels as keys and lists of nodes as values.

When traversing the tree, how do you update the vertical level for the right child?

A vertical level

B vertical level - 1

C vertical level + 1

D vertical level * 2

Right child nodes increase the vertical level by 1.

What traversal method is commonly combined with vertical level tracking for this problem?

A Level order (BFS)

B Inorder

C Postorder

D Preorder

Level order traversal ensures nodes are processed top to bottom.

If two nodes share the same vertical and horizontal level, how is their order decided in output?

A Order of insertion in traversal

B Value size order

C Random order

D Alphabetical order

They appear in the order they are visited during level order traversal.

Explain how vertical levels are assigned and used in Vertical Order Traversal of a binary tree.

Think about how you move left or right from the root.

Describe the role of a queue and a map/dictionary in implementing Vertical Order Traversal.

One helps visit nodes in order, the other helps group them.

Practice

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

easy

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

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

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

D. Nodes are visited in inorder (left, root, 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

   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 A

4. Quick Check:

   Vertical order = group by columns [OK]

Hint: Think columns from left to right, not levels or preorder [OK]

Common Mistakes:

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

2. Which TypeScript data structure is best to store nodes by their vertical column during vertical order traversal?

easy

A. Map

B. Set<number>

C. Array<number>

D. string[]

Solution

1. Step 1: Identify need to map columns to nodes

   We need to store nodes grouped by their column index, so a mapping from column number to list of nodes is required.

2. Step 2: Choose appropriate TypeScript structure

   A Map with keys as column numbers and values as arrays of node values fits perfectly.

3. Final Answer:

   Map<number, number[]> -> Option A

4. Quick Check:

   Map column to node list = Map<number, number[]> [OK]

Hint: Use Map to link column numbers to node arrays [OK]

Common Mistakes:

• Using Set which does not store duplicates or order
• Using plain arrays without column keys
• Using string arrays which don't map columns

3. Given this TypeScript code for vertical order traversal, what is the output?

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val: number) {
    this.val = val;
    this.left = null;
    this.right = null;
  }
}

const root = new TreeNode(1);
root.left = new TreeNode(2);
root.right = new TreeNode(3);
root.left.left = new TreeNode(4);
root.left.right = new TreeNode(5);
root.right.left = new TreeNode(6);
root.right.right = new TreeNode(7);

function verticalOrder(root: TreeNode | null): number[][] {
  if (!root) return [];
  const map = new Map();
  const queue: [TreeNode, number][] = [[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 Array.from(map.entries())
    .sort((a, b) => a[0] - b[0])
    .map(entry => entry[1]);
}

console.log(verticalOrder(root));

medium

A. [[7], [3], [1, 5, 6], [2], [4]]

B. [[4], [2], [1, 5, 6], [3], [7]]

C. [[1], [2, 3], [4, 5, 6, 7]]

D. [[4], [2, 5], [1, 6], [3, 7]]

Solution

1. Step 1: Assign columns to nodes during BFS

   Root (1) at col 0; left child (2) at -1; right child (3) at +1; and so on. Nodes grouped by columns: - -2: 4 - -1: 2 - 0: 1, 5, 6 - 1: 3 - 2: 7

2. Step 2: Sort columns and output node lists

   Sorted columns from left to right: -2, -1, 0, 1, 2. Output arrays accordingly.

3. Final Answer:

   [[4], [2], [1, 5, 6], [3], [7]] -> Option B

4. Quick Check:

   Column grouping matches output D [OK]

Hint: Track column indices during BFS, then sort keys [OK]

Common Mistakes:

• Mixing up column indices for left/right children
• Not sorting columns before output
• Using DFS instead of BFS causing wrong order

4. The following TypeScript code for vertical order traversal has a bug. What is the bug?

function verticalOrder(root: TreeNode | null): number[][] {
  if (!root) return [];
  const map = new Map();
  const queue: [TreeNode, number][] = [[root, 0]];
  while (queue.length) {
    const [node, col] = queue.pop()!;
    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 Array.from(map.entries())
    .sort((a, b) => a[0] - b[0])
    .map(entry => entry[1]);
}

medium

A. map.get(col) returns undefined

B. Missing check for null root

C. Using queue.pop() causes DFS instead of BFS traversal

D. Sorting map entries is incorrect

Solution

1. Step 1: Identify queue operation used

   The code uses queue.pop(), which removes the last element, making it a stack (DFS) behavior.

2. Step 2: Explain impact on vertical order traversal

   Vertical order requires BFS (level order) to maintain correct node order per column. Using pop() breaks this order.

3. Final Answer:

   Using queue.pop() causes DFS instead of BFS traversal -> Option C

4. Quick Check:

   pop() = DFS, need shift() for BFS [OK]

Hint: Use shift() for BFS queue, not pop() [OK]

Common Mistakes:

• Confusing pop() with shift() in queue
• Ignoring traversal order effect on output
• Assuming map.get always returns array without check

5. You want to modify vertical order traversal to sort nodes within the same column by their row (depth) and then by value ascending. Which approach is best?

hard

A. Traverse tree using DFS only and collect nodes in order

B. Use a Set per column to avoid duplicates

C. Sort nodes globally after traversal ignoring columns

D. Store nodes as tuples (row, value) per column and sort before output

Solution

1. Step 1: Understand sorting requirement within columns

   Nodes in the same column must be sorted first by their row (depth), then by value ascending.

2. Step 2: Choose data structure to store row and value

   Store tuples (row, value) for each node in the column map, then sort these tuples before output.

3. Step 3: Explain why other options fail

   Set ignores duplicates and order; DFS alone doesn't guarantee order; global sort ignores column grouping.

4. Final Answer:

   Store nodes as tuples (row, value) per column and sort before output -> Option D

5. Quick Check:

   Sort by row and value per column = store tuples and sort [OK]

Hint: Keep (row, value) pairs per column, then sort [OK]

Common Mistakes:

• Using Set which loses order and duplicates
• Relying on DFS without tracking rows
• Sorting all nodes ignoring column grouping