Vertical Order Traversal of Binary Tree in DSA Go - Build from Scratch

Vertical Order Traversal of Binary Tree

📖 Scenario: You are working on a program that helps visualize a family tree. You want to print the family members in vertical order, from left to right, based on their position in the tree.

🎯 Goal: Build a program that performs vertical order traversal of a binary tree and prints the nodes grouped by their vertical columns.

📋 What You'll Learn

Create a binary tree with exact nodes and structure

Use a map to group nodes by their vertical column index

Traverse the tree using a queue and track horizontal distances

Print the nodes in vertical order from leftmost to rightmost column

💡 Why This Matters

🌍 Real World

Vertical order traversal helps in visualizing hierarchical data like family trees, organizational charts, or layered maps.

💼 Career

Understanding tree traversals and grouping data by position is important for roles in software development, data analysis, and system design.

1

Create the Binary Tree

Create a binary tree with the exact structure: root node with value 1, left child 2, right child 3, left child of 2 is 4, right child of 2 is 5, right child of 3 is 6.

DSA Go

// Define the TreeNode struct and create the binary tree nodes
// Your code here

Hint

Define a struct TreeNode with Val, Left, and Right fields. Then create nodes and link them exactly as described.

2

Setup Map and Queue for Traversal

Create a map called columnTable to group nodes by their column index (int to slice of ints). Create a queue slice of pairs where each pair holds a *TreeNode and its column index as int. Initialize the queue with the root node and column index 0.

DSA Go

package main

import "fmt"

type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

type Pair struct {
    Node  *TreeNode
    Col   int
}

func main() {
    root := &TreeNode{Val: 1}
    root.Left = &TreeNode{Val: 2}
    root.Right = &TreeNode{Val: 3}
    root.Left.Left = &TreeNode{Val: 4}
    root.Left.Right = &TreeNode{Val: 5}
    root.Right.Right = &TreeNode{Val: 6}

    // Create columnTable map and queue slice
    // Your code here
}

Hint

Use make(map[int][]int) to create the map. Define a Pair struct to hold node and column index. Initialize queue with root and column 0.

3

Perform Vertical Order Traversal

Use a for loop to process nodes in queue until empty. For each Pair, append the node's value to columnTable at the node's column. If the node has a left child, enqueue it with column - 1. If the node has a right child, enqueue it with column + 1.

DSA Go

package main

import "fmt"

type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

type Pair struct {
    Node  *TreeNode
    Col   int
}

func main() {
    root := &TreeNode{Val: 1}
    root.Left = &TreeNode{Val: 2}
    root.Right = &TreeNode{Val: 3}
    root.Left.Left = &TreeNode{Val: 4}
    root.Left.Right = &TreeNode{Val: 5}
    root.Right.Right = &TreeNode{Val: 6}

    columnTable := make(map[int][]int)
    queue := []Pair{{Node: root, Col: 0}}

    // Use a for loop to process queue until empty
    // Your code here
}

Hint

Use a loop to dequeue from queue. Append node values to columnTable. Enqueue children with updated column indices.

4

Print Vertical Order Traversal Result

Find the minimum and maximum column indices in columnTable. Use a for loop from minimum to maximum column index to print the nodes in each column as a slice.

DSA Go

package main

import "fmt"

type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

type Pair struct {
    Node  *TreeNode
    Col   int
}

func main() {
    root := &TreeNode{Val: 1}
    root.Left = &TreeNode{Val: 2}
    root.Right = &TreeNode{Val: 3}
    root.Left.Left = &TreeNode{Val: 4}
    root.Left.Right = &TreeNode{Val: 5}
    root.Right.Right = &TreeNode{Val: 6}

    columnTable := make(map[int][]int)
    queue := []Pair{{Node: root, Col: 0}}

    for len(queue) > 0 {
        pair := queue[0]
        queue = queue[1:]
        node := pair.Node
        col := pair.Col

        columnTable[col] = append(columnTable[col], node.Val)

        if node.Left != nil {
            queue = append(queue, Pair{Node: node.Left, Col: col - 1})
        }
        if node.Right != nil {
            queue = append(queue, Pair{Node: node.Right, Col: col + 1})
        }
    }

    // Print nodes in vertical order from min to max column
    // Your code here
}

Hint

Find min and max keys in columnTable. Loop from min to max and print the slices stored at each key.

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