Vertical Order Traversal of Binary Tree in DSA Go - Time & Space Complexity

Time Complexity: Vertical Order Traversal of Binary Tree

O(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.

func verticalOrder(root *TreeNode) [][]int {
    if root == nil {
        return [][]int{}
    }
    type nodeInfo struct {
        node *TreeNode
        col  int
    }
    queue := []nodeInfo{{root, 0}}
    colTable := map[int][]int{}
    minCol, maxCol := 0, 0

    for len(queue) > 0 {
        curr := queue[0]
        queue = queue[1:]
        colTable[curr.col] = append(colTable[curr.col], curr.node.Val)
        if curr.node.Left != nil {
            queue = append(queue, nodeInfo{curr.node.Left, curr.col - 1})
            if curr.col-1 < minCol {
                minCol = curr.col - 1
            }
        }
        if curr.node.Right != nil {
            queue = append(queue, nodeInfo{curr.node.Right, curr.col + 1})
            if curr.col+1 > maxCol {
                maxCol = curr.col + 1
            }
        }
    }

    result := [][]int{}
    for i := minCol; i <= maxCol; i++ {
        result = append(result, colTable[i])
    }
    return result
}
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The main loop processes each node once using a queue.
• How many times: Exactly once per node, so n times for n nodes.

How Execution Grows With Input

As the number of nodes grows, the loop runs once per node, so the work grows linearly.

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

Pattern observation: Doubling the nodes roughly doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time grows directly in proportion to the number of nodes in the tree.

Common Mistake

[X] Wrong: "Because we track columns, the complexity is higher than O(n)."

[OK] Correct: We only visit each node once, and column tracking is done with simple map insertions, so it does not add extra loops over nodes.

Interview Connect

Understanding this helps you explain how breadth-first search can be combined with extra information to solve problems efficiently.

Self-Check

"What if we used depth-first search instead of breadth-first search? How would the time complexity change?"

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