Boundary Traversal of Binary Tree in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new binary tree node with given value.

DSA Go

func newNode(val int) *Node {
    return &Node{
        data: [1],
        left: nil,
        right: nil,
    }
}

Drag options to blanks, or click blank then click option'

Avalue

Bval

Cnode

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name not defined in the function like 'value' or 'node'.

The function parameter is named 'val', so we assign 'val' to the node's data field.

2fill in blank

medium

Complete the code to check if a node is a leaf node (no children).

DSA Go

func isLeaf(node *Node) bool {
    return node != nil && node.left == nil && node.[1] == nil
}

Drag options to blanks, or click blank then click option'

Aparent

Bchild

Cright

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Checking a non-existent field like 'child' or 'parent'.

A leaf node has both left and right children as nil, so we check node.right == nil.

3fill in blank

hard

Fix the error in the code to traverse the left boundary of the tree (excluding leaves).

DSA Go

func addLeftBoundary(root *Node, res *[]int) {
    curr := root.left
    for curr != nil {
        if !isLeaf(curr) {
            *res = append(*res, curr.data)
        }
        if curr.left != nil {
            curr = curr.[1]
        } else {
            curr = curr.right
        }
    }
}

Drag options to blanks, or click blank then click option'

Aleft

Bright

Cparent

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using curr.right instead of curr.left when left child exists.

To traverse the left boundary, we move down the left child if it exists.

4fill in blank

hard

Fill both blanks to add the right boundary nodes in reverse order (excluding leaves).

DSA Go

func addRightBoundary(root *Node, res *[]int) {
    curr := root.[1]
    var tmp []int
    for curr != nil {
        if !isLeaf(curr) {
            tmp = append(tmp, curr.data)
        }
        if curr.[2] != nil {
            curr = curr.[2]
        } else {
            curr = curr.left
        }
    }
    for i := len(tmp) - 1; i >= 0; i-- {
        *res = append(*res, tmp[i])
    }
}

Drag options to blanks, or click blank then click option'

Aright

Bleft

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from root.left or moving down left children for right boundary.

Start from root.right and move down right children to collect right boundary nodes.

5fill in blank

hard

Fill all three blanks to collect all leaf nodes in left-to-right order.

DSA Go

func addLeaves(root *Node, res *[]int) {
    if root == nil {
        return
    }
    if isLeaf(root) {
        *res = append(*res, root.[1])
        return
    }
    addLeaves(root.[2], res)
    addLeaves(root.[3], res)
}

Drag options to blanks, or click blank then click option'

Adata

Bleft

Cright

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Appending wrong field like 'value' or traversing right child before left.

Append the data of leaf nodes and recursively traverse left then right children.

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