Vertical Order Traversal of Binary Tree in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a struct for a binary tree node.

DSA Go

type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *[1]
}

Drag options to blanks, or click blank then click option'

Astring

BTreeNode

Cint

DNode

Attempts:

3 left

💡 Hint

Common Mistakes

Using int or string instead of *TreeNode for child pointers.

Forgetting the pointer symbol * before TreeNode.

The Right child of a binary tree node is a pointer to another TreeNode struct.

2fill in blank

medium

Complete the code to initialize a map that stores vertical order nodes by their horizontal distance.

DSA Go

verticalMap := make(map[int][]int)

// Add root value at horizontal distance 0
verticalMap[[1]] = append(verticalMap[0], root.Val)

Drag options to blanks, or click blank then click option'

A0

B-1

C1

Droot.Val

Attempts:

3 left

💡 Hint

Common Mistakes

Using root.Val as the key instead of 0.

Using 1 or -1 which represent child nodes, not root.

The root node is at horizontal distance 0, so we use 0 as the key in the map.

3fill in blank

hard

Fix the error in the recursive call to traverse the left subtree with updated horizontal distance.

DSA Go

func traverse(node *TreeNode, hd int, verticalMap map[int][]int) {
    if node == nil {
        return
    }
    verticalMap[hd] = append(verticalMap[hd], node.Val)
    traverse(node.Left, hd [1] 1, verticalMap)
    traverse(node.Right, hd + 1, verticalMap)
}

Drag options to blanks, or click blank then click option'

A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '-' for left child horizontal distance.

Using multiplication or division operators which are incorrect here.

For the left child, horizontal distance decreases by 1, so use hd - 1.

4fill in blank

hard

Fill both blanks to sort the keys of the verticalMap and collect values in order.

DSA Go

var keys []int
for k := range verticalMap {
    keys = append(keys, [1])
}
sort.[2](keys)

var result [][]int
for _, k := range keys {
    result = append(result, verticalMap[k])
}

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Ak

Bkeys

CInts

DSort

Attempts:

3 left

💡 Hint

Common Mistakes

Appending keys instead of k in the loop.

Using sort.Sort or sort.Sort(keys) which requires a different type.

Using capitalized Sort instead of Ints.

We append each key k to keys slice, then sort keys using sort.Ints to get ascending order.

5fill in blank

hard

Fill all three blanks to complete the vertical order traversal function returning the result.

DSA Go

func verticalOrder(root *TreeNode) [][]int {
    verticalMap := make(map[int][]int)
    traverse(root, 0, verticalMap)

    var keys []int
    for k := range verticalMap {
        keys = append(keys, [1])
    }
    sort.[2](keys)

    var result [][]int
    for _, k := range keys {
        result = append(result, verticalMap[[3]])
    }
    return result
}

Drag options to blanks, or click blank then click option'

Ak

BInts

DSort

Attempts:

3 left

💡 Hint

Common Mistakes

Using Sort instead of Ints for sorting.

Using wrong variable names instead of k.

Forgetting to return the result slice.

We append each key k, sort keys with sort.Ints, and append verticalMap[k] to result.

Practice

1. What does vertical order traversal of a binary tree primarily group nodes by?

easy

A. Their parent node

B. Their depth level in the tree

C. Their value magnitude

D. Their horizontal distance from the root

Solution

1. Step 1: Understand vertical order traversal concept

   Vertical order traversal groups nodes based on their horizontal distance from the root node, also called the vertical column.

2. Step 2: Differentiate from other traversals

   Unlike level order which groups by depth, vertical order groups nodes that align vertically when the tree is viewed from top.

3. Final Answer:

   Their horizontal distance from the root -> Option D

4. Quick Check:

   Vertical order = horizontal distance grouping [OK]

Hint: Remember vertical means grouping by horizontal distance [OK]

Common Mistakes:

• Confusing vertical order with level order traversal
• Grouping by node values instead of position
• Ignoring horizontal distance concept

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

easy

A. map[string]int

B. []int

C. map[int][]int

D. [][]int

Solution

1. Step 1: Identify grouping requirement

   We need to group nodes by their vertical column index, which is an integer key.

2. Step 2: Choose suitable Go data structure

   A map with integer keys and slice of integers as values (map[int][]int) allows storing multiple node values per vertical column.

3. Final Answer:

   map[int][]int -> Option C

4. Quick Check:

   Use map[int][]int for vertical groups [OK]

Hint: Use map with int keys and slice values for grouping [OK]

Common Mistakes:

• Using simple slices without keys
• Using map with string keys unnecessarily
• Choosing 2D slices without keys

3. Given the binary tree:

    1
   / \
  2   3
 /     \
4       5

What is the vertical order traversal output?

medium

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

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

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

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

Solution

1. Step 1: Assign horizontal distances

   Root 1 is at 0, left child 2 at -1, left-left 4 at -2, right child 3 at +1, right-right 5 at +2.

2. Step 2: Group nodes by horizontal distance

   At -2: [4], -1: [2], 0: [1], +1: [3], +2: [5].

3. Final Answer:

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

4. Quick Check:

   Group by horizontal distance = [[4], [2], [1], [3, 5]] [OK]

Hint: Label columns from left (-) to right (+) then group nodes [OK]

Common Mistakes:

• Mixing node order inside same vertical column
• Incorrect horizontal distance assignment
• Ignoring multiple nodes in same vertical column

4. Consider this Go snippet for vertical order traversal:

func verticalOrder(root *Node) map[int][]int {
    result := make(map[int][]int)
    var dfs func(node *Node, hd int)
    dfs = func(node *Node, hd int) {
        if node == nil {
            return
        }
        result[hd] = append(result[hd], node.Val)
        dfs(node.Left, hd-1)
        dfs(node.Right, hd+1)
    }
    dfs(root, 0)
    return result
}

What is the main issue with this code?

medium

A. It uses DFS which may not preserve top-to-bottom order

B. It does not handle nil root input

C. It uses map instead of slice for result

D. It increments horizontal distance incorrectly

Solution

1. Step 1: Analyze traversal method

   The code uses DFS which explores depth first, so nodes deeper may be added before shallower nodes in same vertical column.

2. Step 2: Understand vertical order requirement

   Vertical order traversal requires nodes to be output top-to-bottom per vertical column, which BFS preserves but DFS may not.

3. Final Answer:

   It uses DFS which may not preserve top-to-bottom order -> Option A

4. Quick Check:

   DFS can misorder vertical nodes top-to-bottom [OK]

Hint: Use BFS, not DFS, to keep top-to-bottom order [OK]

Common Mistakes:

• Assuming DFS preserves vertical order top-to-bottom
• Ignoring traversal order impact on output
• Thinking map usage causes order issues

5. You want to print the vertical order traversal of a binary tree sorted by vertical column from left to right and nodes top to bottom within each column. Which approach is best?

hard

A. Use BFS with a queue storing nodes and their horizontal distance, then sort keys

B. Use DFS with recursion and store nodes in map without sorting

C. Use inorder traversal and print nodes as visited

D. Use a stack to store nodes and print after traversal

Solution

1. Step 1: Choose traversal preserving top-to-bottom order

   BFS visits nodes level by level, ensuring top nodes are processed before lower nodes in same vertical column.

2. Step 2: Store nodes with horizontal distance and sort keys

   Use a queue to track nodes and their horizontal distance, store in map[int][]int, then output sorted by keys for left to right order.

3. Final Answer:

   Use BFS with a queue storing nodes and their horizontal distance, then sort keys -> Option A

4. Quick Check:

   BFS + sorting keys = correct vertical order traversal [OK]

Hint: BFS + map + sort keys for correct vertical order [OK]

Common Mistakes:

• Using DFS which breaks top-to-bottom order
• Not sorting map keys before printing
• Using stack which reverses order