Zigzag Level Order Traversal in DSA Go - Time & Space Complexity

Time Complexity: Zigzag Level Order Traversal

O(n)

Understanding Time Complexity

We want to understand how the time needed to traverse a tree in zigzag order changes as the tree grows.

How does the number of steps grow when the tree has more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func zigzagLevelOrder(root *TreeNode) [][]int {
    if root == nil {
        return [][]int{}
    }
    var result [][]int
    queue := []*TreeNode{root}
    leftToRight := true

    for len(queue) > 0 {
        levelSize := len(queue)
        level := make([]int, levelSize)
        for i := 0; i < levelSize; i++ {
            node := queue[0]
            queue = queue[1:]
            index := i
            if !leftToRight {
                index = levelSize - 1 - i
            }
            level[index] = node.Val
            if node.Left != nil {
                queue = append(queue, node.Left)
            }
            if node.Right != nil {
                queue = append(queue, node.Right)
            }
        }
        result = append(result, level)
        leftToRight = !leftToRight
    }
    return result
}
    

This code visits each node of a binary tree level by level, alternating the order of values between left-to-right and right-to-left.

Identify Repeating Operations

• Primary operation: The outer loop runs once per tree level, and the inner loop visits every node in that level.
• How many times: Each node is visited exactly once in total during the traversal.

How Execution Grows With Input

As the number of nodes (n) grows, the code visits each node once, so the total steps grow roughly in direct proportion to n.

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

Pattern observation: The work grows linearly as the tree size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to complete the traversal grows directly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "Because we have nested loops, the time complexity must be O(n²)."

[OK] Correct: The inner loop runs only for nodes at each level, and all nodes together are visited once total, so the total work is still proportional to n, not n squared.

Interview Connect

Understanding this traversal's time complexity helps you explain how tree traversals scale, a common topic in interviews. It shows you can analyze loops carefully, not just count nesting.

Self-Check

"What if we used a recursive approach instead of a queue? How would the time complexity change?"

Practice

1. What does the Zigzag Level Order Traversal of a binary tree do differently compared to a normal level order traversal?

easy

A. It alternates the direction of traversal at each level, switching between left-to-right and right-to-left.

B. It only traverses the left subtree of the tree.

C. It visits nodes in a random order at each level.

D. It traverses nodes only in right-to-left order for all levels.

Solution

1. Step 1: Understand normal level order traversal

   Normal level order traversal visits nodes level by level from left to right.

2. Step 2: Identify zigzag difference

   Zigzag traversal changes direction every level: first left-to-right, then right-to-left, and so on.

3. Final Answer:

   It alternates the direction of traversal at each level, switching between left-to-right and right-to-left. -> Option A

4. Quick Check:

   Zigzag = alternate directions [OK]

Hint: Remember zigzag means direction flips each level [OK]

Common Mistakes:

• Thinking traversal is always left-to-right
• Confusing zigzag with random order
• Ignoring direction change at each level

2. Which of the following Go code snippets correctly initializes a queue for level order traversal in Zigzag Level Order Traversal?

easy

A. queue := make([]int, 0)

B. queue := []int{root}

C. queue := map[int]*TreeNode{}

D. queue := []*TreeNode{root}

Solution

1. Step 1: Identify the queue type needed

   Queue should hold pointers to TreeNode structs, so []*TreeNode is correct.

2. Step 2: Check each option's correctness

   queue := []*TreeNode{root} uses []*TreeNode with root node, correct for traversal. Options A and B use int slices, wrong type. queue := map[int]*TreeNode{} uses map, not queue.

3. Final Answer:

   queue := []*TreeNode{root} -> Option D

4. Quick Check:

   Queue holds TreeNode pointers = queue := []*TreeNode{root} [OK]

Hint: Queue must hold TreeNode pointers, not ints or maps [OK]

Common Mistakes:

• Using int slices instead of TreeNode pointers
• Using map instead of slice for queue
• Not initializing queue with root node

3. Given the binary tree:

    1
   / \
  2   3
 / \   \
4   5   6

What is the output of Zigzag Level Order Traversal?

medium

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

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

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

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

Solution

1. Step 1: Traverse level 0 and 1 with direction

   Level 0: left-to-right -> [1]. Level 1: right-to-left -> [3, 2].

2. Step 2: Traverse level 2 left-to-right

   Level 2 nodes: 4, 5, 6 from left to right -> [4, 5, 6].

3. Final Answer:

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

4. Quick Check:

   Zigzag levels alternate directions = [[1], [3, 2], [4, 5, 6]] [OK]

Hint: Alternate direction each level, left-to-right then right-to-left [OK]

Common Mistakes:

• Not reversing nodes on odd levels
• Reversing nodes on even levels incorrectly
• Mixing node order at last level

4. Consider this Go code snippet for Zigzag Level Order Traversal:

for len(queue) > 0 {
    levelSize := len(queue)
    var level []int
    for i := 0; i < levelSize; i++ {
        node := queue[0]
        queue = queue[1:]
        level = append(level, node.Val)
        if node.Left != nil {
            queue = append(queue, node.Left)
        }
        if node.Right != nil {
            queue = append(queue, node.Right)
        }
    }
    if leftToRight {
        result = append(result, level)
    } else {
        reverse(level)
        result = append(result, level)
    }
    leftToRight = !leftToRight
}

What is the main bug in this code?

medium

A. The reverse function is called but not defined or implemented.

B. The queue is not updated correctly after popping the first element.

C. The variable leftToRight is never initialized before use.

D. The level slice is appended before processing child nodes.

Solution

1. Step 1: Check queue update correctness

   Queue is updated by slicing off first element correctly, no bug here.

2. Step 2: Verify reverse function usage

   reverse(level) is called but no definition or implementation is shown, causing compile error.

3. Final Answer:

   The reverse function is called but not defined or implemented. -> Option A

4. Quick Check:

   Undefined reverse function causes error [OK]

Hint: Ensure all helper functions like reverse are implemented [OK]

Common Mistakes:

• Assuming reverse is built-in in Go
• Ignoring initialization of leftToRight (usually done before loop)
• Misunderstanding queue slicing syntax

5. You want to implement Zigzag Level Order Traversal but optimize space by avoiding reversing slices. Which approach is best?

hard

A. Use a queue and reverse the slice after each level as usual.

B. Use two stacks to alternate pushing nodes in correct order per level.

C. Use recursion with a global list and reverse at the end.

D. Traverse all nodes left-to-right and then reverse the entire result.

Solution

1. Step 1: Understand reversing cost

   Reversing slices after each level adds extra time and space overhead.

2. Step 2: Use two stacks for order control

   Two stacks let you push child nodes in order that naturally produces zigzag traversal without reversing.

3. Final Answer:

   Use two stacks to alternate pushing nodes in correct order per level. -> Option B

4. Quick Check:

   Two stacks avoid reversing slices [OK]

Hint: Two stacks method avoids reversing slices [OK]

Common Mistakes:

• Relying on reversing slices every level
• Using recursion without order control
• Reversing entire result instead of per level