Zigzag Level Order Traversal in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Zigzag Level Order Traversal on a Binary Tree

What is the output of the zigzag level order traversal for the given binary tree?

DSA Go

package main

import "fmt"

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

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

    for len(currentLevel) > 0 {
        var levelVals []int
        var nextLevel []*TreeNode

        for _, node := range currentLevel {
            levelVals = append(levelVals, node.Val)
        }

        if !leftToRight {
            for i, j := 0, len(levelVals)-1; i < j; i, j = i+1, j-1 {
                levelVals[i], levelVals[j] = levelVals[j], levelVals[i]
            }
        }

        for _, node := range currentLevel {
            if node.Left != nil {
                nextLevel = append(nextLevel, node.Left)
            }
            if node.Right != nil {
                nextLevel = append(nextLevel, node.Right)
            }
        }

        result = append(result, levelVals)
        currentLevel = nextLevel
        leftToRight = !leftToRight
    }

    return result
}

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}

    result := zigzagLevelOrder(root)
    fmt.Println(result)
}

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

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

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

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

💡 Hint

Remember that zigzag means alternating the order of nodes at each level.

The traversal starts left to right at level 0, so [1]. Then it goes right to left at level 1, so [3, 2]. Finally, it goes left to right at level 2, so [4, 5, 6].

🧠 Conceptual

intermediate

Understanding Zigzag Level Order Traversal Direction

In zigzag level order traversal, what determines the order of nodes at each level?

A The level number: even levels left to right, odd levels right to left

B The number of children a node has

C The value of the node: even values left to right, odd values right to left

D The height of the tree

💡 Hint

Think about how the traversal direction changes as you go down levels.

Zigzag traversal alternates direction by level number: even levels are traversed left to right, odd levels right to left.

🔧 Debug

advanced

Identify the Bug in Zigzag Traversal Code

What error will this code produce when run, and why? func zigzagLevelOrder(root *TreeNode) [][]int { if root == nil { return [][]int{} } var result [][]int var currentLevel []*TreeNode currentLevel = append(currentLevel, root) leftToRight := true for len(currentLevel) > 0 { var levelVals []int var nextLevel []*TreeNode for i := len(currentLevel) - 1; i >= 0; i-- { node := currentLevel[i] levelVals = append(levelVals, node.Val) if leftToRight { if node.Left != nil { nextLevel = append(nextLevel, node.Left) } if node.Right != nil { nextLevel = append(nextLevel, node.Right) } } else { if node.Right != nil { nextLevel = append(nextLevel, node.Right) } if node.Left != nil { nextLevel = append(nextLevel, node.Left) } } } result = append(result, levelVals) currentLevel = nextLevel leftToRight = !leftToRight } return result }

A The code will panic due to nil pointer dereference

B The code will cause an infinite loop

C The code will run correctly and produce the expected zigzag output

D The traversal order of nodes in nextLevel is incorrect, causing wrong zigzag output

💡 Hint

Check how nextLevel nodes are added and in what order.

The order of adding children to nextLevel is wrong because appending in this order while iterating backwards causes the next level to be in incorrect order, breaking the zigzag pattern.

🚀 Application

advanced

Number of Levels in Zigzag Traversal Output

Given a binary tree with 7 nodes arranged as a complete binary tree of height 3, how many levels will the zigzag level order traversal output contain?

A 3

B 7

C 4

D 2

💡 Hint

Think about how many levels a complete binary tree of height 3 has.

A complete binary tree of height 3 has exactly 3 levels, so the traversal output will have 3 lists.

❓ Predict Output

expert

Output of Zigzag Traversal on Unbalanced Tree

What is the output of the zigzag level order traversal for this unbalanced binary tree? Tree structure: - Root: 10 - Root.Left: 5 - Root.Left.Left: 3 - Root.Left.Left.Left: 1 - Root.Right: 15 - Root.Right.Right: 20

DSA Go

package main

import "fmt"

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

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

    for len(currentLevel) > 0 {
        var levelVals []int
        var nextLevel []*TreeNode

        for _, node := range currentLevel {
            levelVals = append(levelVals, node.Val)
        }

        if !leftToRight {
            for i, j := 0, len(levelVals)-1; i < j; i, j = i+1, j-1 {
                levelVals[i], levelVals[j] = levelVals[j], levelVals[i]
            }
        }

        for _, node := range currentLevel {
            if node.Left != nil {
                nextLevel = append(nextLevel, node.Left)
            }
            if node.Right != nil {
                nextLevel = append(nextLevel, node.Right)
            }
        }

        result = append(result, levelVals)
        currentLevel = nextLevel
        leftToRight = !leftToRight
    }

    return result
}

func main() {
    root := &TreeNode{Val: 10}
    root.Left = &TreeNode{Val: 5}
    root.Left.Left = &TreeNode{Val: 3}
    root.Left.Left.Left = &TreeNode{Val: 1}
    root.Right = &TreeNode{Val: 15}
    root.Right.Right = &TreeNode{Val: 20}

    result := zigzagLevelOrder(root)
    fmt.Println(result)
}

A [[10], [15, 5], [20, 3], [1]]

B [[10], [5, 15], [3, 20], [1]]

C [[10], [15, 5], [3, 20], [1]]

D [[10], [5, 15], [20, 3], [1]]

💡 Hint

Remember to alternate the order of nodes at each level, even if the tree is unbalanced.

The traversal starts left to right at level 0: [10]. Then right to left at level 1: [15, 5]. Then left to right at level 2: [3, 20]. Finally right to left at level 3: [1].

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