Boundary Traversal of Binary Tree in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Boundary Traversal on a Simple Binary Tree

What is the output of the boundary traversal for the given binary tree?

DSA Go

package main

import "fmt"

type Node struct {
    data  int
    left  *Node
    right *Node
}

func boundaryTraversal(root *Node) []int {
    if root == nil {
        return []int{}
    }
    var res []int
    res = append(res, root.data)

    // Left boundary (excluding leaf)
    curr := root.left
    for curr != nil {
        if curr.left != nil || curr.right != nil {
            res = append(res, curr.data)
        }
        if curr.left != nil {
            curr = curr.left
        } else {
            curr = curr.right
        }
    }

    // Leaves
    var addLeaves func(*Node)
    addLeaves = func(node *Node) {
        if node == nil {
            return
        }
        if node.left == nil && node.right == nil {
            res = append(res, node.data)
            return
        }
        addLeaves(node.left)
        addLeaves(node.right)
    }
    addLeaves(root.left)
    addLeaves(root.right)

    // Right boundary (excluding leaf) in reverse
    var stack []int
    curr = root.right
    for curr != nil {
        if curr.left != nil || curr.right != nil {
            stack = append(stack, curr.data)
        }
        if curr.right != nil {
            curr = curr.right
        } else {
            curr = curr.left
        }
    }
    for i := len(stack) - 1; i >= 0; i-- {
        res = append(res, stack[i])
    }

    return res
}

func main() {
    root := &Node{1, nil, nil}
    root.left = &Node{2, nil, nil}
    root.right = &Node{3, nil, nil}
    root.left.left = &Node{4, nil, nil}
    root.left.right = &Node{5, nil, nil}
    root.right.left = &Node{6, nil, nil}
    root.right.right = &Node{7, nil, nil}

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

A [1, 2, 4, 5, 6, 7, 3]

B [1, 2, 4, 5, 7, 6, 3]

C [1, 2, 4, 5, 6, 3, 7]

D [1, 4, 5, 6, 7, 2, 3]

💡 Hint

Remember to include left boundary (excluding leaves), then leaves, then right boundary in reverse order.

The boundary traversal starts with root, then left boundary excluding leaves (2), then leaves (4,5,6,7), then right boundary excluding leaves in reverse (3). So the output is [1, 2, 4, 5, 6, 7, 3].

🧠 Conceptual

intermediate

Understanding Boundary Traversal Components

Which of the following correctly describes the order of nodes visited in a boundary traversal of a binary tree?

A Root node, all leaf nodes from left to right, left boundary excluding leaves, right boundary excluding leaves in reverse

B All leaf nodes from left to right, root node, left boundary excluding leaves, right boundary excluding leaves in reverse

C Root node, left boundary excluding leaves, all leaf nodes from left to right, right boundary excluding leaves in reverse

D Left boundary excluding leaves, root node, right boundary excluding leaves in reverse, all leaf nodes from left to right

💡 Hint

Think about the natural boundary around the tree starting from the root.

Boundary traversal visits the root first, then the left boundary excluding leaves, then all leaves from left to right, and finally the right boundary excluding leaves in reverse order.

🔧 Debug

advanced

Identify the Bug in Boundary Traversal Code

Given the following snippet from a boundary traversal function, what error will it cause if root == nil? func boundaryTraversal(root *Node) []int { var res []int res = append(res, root.data) curr := root.left for curr != nil { if curr.left != nil || curr.right != nil { res = append(res, curr.data) } if curr.left != nil { curr = curr.left } else { curr = curr.right } } // Leaves and right boundary code omitted for brevity return res }

A Compilation error due to missing return statement

B Panic due to nil pointer dereference when accessing root.data if root is nil

C No error, returns slice with only root data

D Infinite loop because curr never becomes nil

💡 Hint

Check if root is nil before accessing its fields.

If root is nil, accessing root.data causes a runtime panic due to nil pointer dereference.

❓ Predict Output

advanced

Boundary Traversal Output for Skewed Tree

What is the output of the boundary traversal for the following left-skewed binary tree? Tree structure: 1 | 2 | 3 | 4 Code snippet: func main() { root := &Node{1, nil, nil} root.left = &Node{2, nil, nil} root.left.left = &Node{3, nil, nil} root.left.left.left = &Node{4, nil, nil} result := boundaryTraversal(root) fmt.Println(result) }

A [1, 2, 4, 3]

B [1, 4, 3, 2]

C [4, 3, 2, 1]

D [1, 2, 3, 4]

💡 Hint

In a skewed tree, left boundary and leaves overlap.

The boundary traversal visits root (1), left boundary excluding leaves (2,3), leaves (4), and no right boundary. So output is [1, 2, 3, 4].

🚀 Application

expert

Number of Nodes in Boundary Traversal Result

Given a perfect binary tree of height 2 (levels 0 to 2), how many nodes will be in the boundary traversal output?

A 7

B 8

C 9

D 10

💡 Hint

Count root (1), left boundary excluding leaves (1), leaves (4), and right boundary excluding leaves (1).

A perfect binary tree of height 2 (levels 0 to 2) has 7 nodes, all of which are on the boundary: root (1), left boundary excluding leaves (1 node at level 1 left), leaves (4 at level 2), right boundary excluding leaves (1 node at level 1 right). Total: 7.

Practice

1. What is the main purpose of Boundary Traversal in a binary tree?

easy

A. To print nodes level by level

B. To print all leaf nodes only

C. To print nodes in in-order traversal

D. To print all nodes on the boundary of the tree in an anti-clockwise direction

Solution

1. Step 1: Understand boundary traversal definition

   Boundary traversal prints nodes on the edges of the tree, including left boundary, leaves, and right boundary.

2. Step 2: Compare with other traversals

   Unlike leaf-only or level order, boundary traversal covers all edges in anti-clockwise order.

3. Final Answer:

   To print all nodes on the boundary of the tree in an anti-clockwise direction -> Option D

4. Quick Check:

   Boundary traversal = print edges anti-clockwise [OK]

Hint: Boundary traversal prints edges anti-clockwise [OK]

Common Mistakes:

• Confusing boundary traversal with leaf node traversal
• Thinking it prints nodes level-wise
• Mixing it with in-order traversal

2. Which of the following Go function signatures correctly represents a function to perform boundary traversal of a binary tree node?

easy

A. func boundaryTraversal(root Node) int

B. func boundaryTraversal(root *Node) []int

C. func boundaryTraversal(root *Node) int

D. func boundaryTraversal(root Node) []Node

Solution

1. Step 1: Understand function input and output

   The function should take a pointer to the root node and return a slice of integers representing node values.

2. Step 2: Check each option

   func boundaryTraversal(root *Node) []int matches input as pointer and output as slice of int, suitable for traversal results.

3. Final Answer:

   func boundaryTraversal(root *Node) []int -> Option B

4. Quick Check:

   Pointer input + slice output = correct signature [OK]

Hint: Use pointer receiver and slice return for traversal [OK]

Common Mistakes:

• Passing node by value instead of pointer
• Returning single int instead of slice
• Returning slice of nodes instead of values

3. Given the following binary tree and Go code snippet, what is the output of the boundary traversal? Tree structure: ``` 1 / \ 2 3 / \ \ 4 5 6 / \ / 7 8 9 ``` Go code snippet:

func main() {
    root := &Node{1, nil, nil}
    root.Left = &Node{2, nil, nil}
    root.Right = &Node{3, nil, nil}
    root.Left.Left = &Node{4, nil, nil}
    root.Left.Right = &Node{5, nil, nil}
    root.Right.Right = &Node{6, nil, nil}
    root.Left.Right.Left = &Node{7, nil, nil}
    root.Left.Right.Right = &Node{8, nil, nil}
    root.Right.Right.Left = &Node{9, nil, nil}

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

medium

A. [1 2 4 7 8 9 6 3]

B. [1 2 4 7 8 5 9 6 3]

C. [1 2 4 7 8 9 6]

D. [1 2 4 5 7 8 9 6 3]

Solution

1. Step 1: Identify boundary nodes in anti-clockwise order

   Left boundary: 1, 2, 4 (excluding leaves repeated), Leaves: 4,7,8,9,6, Right boundary: 3 (excluding leaf 6 as it's already included)

2. Step 2: Combine nodes without duplicates

   Boundary traversal = 1 (root), 2 (left boundary), 4 (leaf), 7 (leaf), 8 (leaf), 9 (leaf), 6 (leaf), 3 (right boundary)

3. Final Answer:

   [1 2 4 7 8 9 6 3] -> Option A

4. Quick Check:

   Boundary nodes combined = [1 2 4 7 8 9 6 3] [OK]

Hint: List left boundary, leaves, then right boundary reversed [OK]

Common Mistakes:

• Including internal nodes like 5 in boundary
• Repeating leaf nodes in boundary and leaves
• Wrong order of right boundary nodes

4. The following Go function for boundary traversal has a bug. What is the main issue?

func boundaryTraversal(root *Node) []int {
    if root == nil {
        return []int{}
    }
    result := []int{root.Val}
    addLeftBoundary(root.Left, &result)
    addLeaves(root, &result)
    addRightBoundary(root.Right, &result)
    return result
}

func addLeftBoundary(node *Node, res *[]int) {
    for node != nil {
        if node.Left != nil || node.Right != nil {
            *res = append(*res, node.Val)
        }
        if node.Left != nil {
            node = node.Left
        } else {
            node = node.Right
        }
    }
}

func addLeaves(node *Node, res *[]int) {
    if node == nil {
        return
    }
    if node.Left == nil && node.Right == nil {
        *res = append(*res, node.Val)
        return
    }
    addLeaves(node.Left, res)
    addLeaves(node.Right, res)
}

func addRightBoundary(node *Node, res *[]int) {
    temp := []int{}
    for node != nil {
        if node.Left != nil || node.Right != nil {
            temp = append(temp, node.Val)
        }
        if node.Right != nil {
            node = node.Right
        } else {
            node = node.Left
        }
    }
    for i := len(temp) - 1; i >= 0; i-- {
        *res = append(*res, temp[i])
    }
}

medium

A. Leaves are added twice because addLeaves includes left and right boundary leaves

B. Right boundary nodes are appended in wrong order

C. Left boundary includes leaf nodes which should be excluded

D. Root node is not added to the result

Solution

1. Step 1: Analyze addLeftBoundary function

   It adds nodes if they have left or right child, but does not exclude leaf nodes properly, so leaf nodes may be added twice.

2. Step 2: Check leaf node addition

   Leaves are added separately by addLeaves, so left boundary should exclude leaves to avoid duplicates.

3. Final Answer:

   Left boundary includes leaf nodes which should be excluded -> Option C

4. Quick Check:

   Left boundary must exclude leaves [OK]

Hint: Exclude leaves from left boundary to avoid duplicates [OK]

Common Mistakes:

• Not excluding leaves in left boundary
• Appending right boundary in forward order
• Forgetting to add root node

5. You want to modify the boundary traversal to print nodes in clockwise order starting from root. Which change is correct to achieve this?

hard

A. Traverse right boundary top-down, then leaves right-to-left, then left boundary bottom-up

B. Traverse left boundary top-down, then leaves left-to-right, then right boundary bottom-up

C. Traverse leaves first, then left boundary, then right boundary

D. Traverse right boundary bottom-up, then leaves left-to-right, then left boundary top-down

Solution

1. Step 1: Understand clockwise boundary traversal order

   Clockwise order starts from root, then right boundary top-down, leaves right-to-left, and left boundary bottom-up.

2. Step 2: Match options with traversal order

   Traverse right boundary top-down, then leaves right-to-left, then left boundary bottom-up matches this order exactly, others do not.

3. Final Answer:

   Traverse right boundary top-down, then leaves right-to-left, then left boundary bottom-up -> Option A

4. Quick Check:

   Clockwise = root, right boundary, leaves, left boundary [OK]

Hint: Clockwise = root + right boundary + leaves + left boundary [OK]

Common Mistakes:

• Using left boundary first for clockwise
• Not reversing leaf order for clockwise
• Mixing top-down and bottom-up incorrectly