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

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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.
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beginner
How do we assign horizontal distances (HD) to nodes in Vertical Order Traversal?
The root node is assigned HD = 0. For any node, its left child has HD = parent's HD - 1, and its right child has HD = parent's HD + 1.
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intermediate
Why do we use a queue and a map in Vertical Order Traversal?
A queue helps in level order traversal to process nodes level by level. A map stores nodes grouped by their horizontal distance for final vertical order output.
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intermediate
What data structure is commonly used to keep track of nodes and their horizontal distances during traversal?
A queue of pairs (node, horizontal distance) is used to process nodes in order and track their vertical positions.
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advanced
How do we handle nodes that share the same horizontal distance and level in Vertical Order Traversal?
They are printed in the order they appear in the level order traversal, usually from left to right.
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In Vertical Order Traversal, what horizontal distance is assigned to the root node?
A1
B-1
C0
DDepends on tree height
Which data structure is best to store nodes grouped by their horizontal distance?
AMap or Dictionary
BArray
CStack
DLinked List
What traversal method is combined with horizontal distance tracking for Vertical Order Traversal?
APostorder
BInorder
CPreorder
DLevel order (BFS)
If a node is at horizontal distance 2, where is it located relative to the root?
ATwo steps right
BTwo steps left
CAt root
DCannot determine
How do we print the final vertical order after traversal?
AFrom highest to lowest horizontal distance
BFrom lowest to highest horizontal distance
CRandom order
DOnly nodes at horizontal distance 0
Explain step-by-step how to perform Vertical Order Traversal on a binary tree.
Think about how to combine level order traversal with horizontal distance tracking.
You got /5 concepts.
    Describe how horizontal distances change when moving to left and right children in a binary tree.
    Consider the root as the center point.
    You got /3 concepts.

      Practice

      (1/5)
      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
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