Tree Terminology Root Leaf Height Depth Level in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create the root node of a tree.

DSA Go

root := &Node{Value: 10, [1]: nil, Left: nil, Right: nil}

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

AParent

BChild

CRoot

DLeaf

Attempts:

3 left

💡 Hint

Common Mistakes

Setting Parent to a non-nil value for root node.

The root node has no parent, so its Parent field is nil.

2fill in blank

medium

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

DSA Go

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

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

AParent

BRight

CRoot

DValue

Attempts:

3 left

💡 Hint

Common Mistakes

Checking Parent instead of Right child.

A leaf node has no left or right children, so both Left and Right are nil.

3fill in blank

hard

Fix the error in the code to calculate the height of a node.

DSA Go

func height(node *Node) int {
    if node == nil {
        return -1
    }
    leftHeight := height(node.Left)
    rightHeight := height(node.[1])
    if leftHeight > rightHeight {
        return leftHeight + 1
    }
    return rightHeight + 1
}

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

AParent

BRoot

CValue

DRight

Attempts:

3 left

💡 Hint

Common Mistakes

Using Parent or Value instead of Right child.

To find height, we check both Left and Right children recursively.

4fill in blank

hard

Fill both blanks to compute the depth of a node in the tree.

DSA Go

func depth(node *Node) int {
    if node.[1] == nil {
        return 0
    }
    return 1 + depth(node.[2])
}

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

AParent

BLeft

CRight

DValue

Attempts:

3 left

💡 Hint

Common Mistakes

Using Left or Right child pointers instead of Parent.

Depth is the number of edges from the node to the root, so we move up using Parent.

5fill in blank

hard

Fill all three blanks to create a map of node levels in the tree.

DSA Go

func assignLevels(node *Node, level int, levels map[*Node]int) {
    if node == nil {
        return
    }
    levels[node] = [1]
    assignLevels(node.[2], level [3], levels)
    assignLevels(node.Right, level+1, levels)
}

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

Alevel

B+ 1

C- 1

Dlevel + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect level increments or decrements.

We assign the current level to the node, then recurse on children with level + 1.

Practice

1. In a tree data structure, which node is called the root?

easy

A. The topmost node with no parent

B. A node with no children

C. Any node at the lowest level

D. A node with exactly two children

Solution

1. Step 1: Understand the root node definition

   The root node is the starting point of a tree and has no parent node above it.

2. Step 2: Differentiate root from other nodes

   Leaf nodes have no children, but root specifically has no parent. Lowest level nodes are leaves or internal nodes, not necessarily root.

3. Final Answer:

   The topmost node with no parent -> Option A

4. Quick Check:

   Root = topmost node with no parent [OK]

Hint: Root is always the top node without a parent [OK]

Common Mistakes:

• Confusing root with leaf node
• Thinking root must have children
• Assuming root is at the lowest level

2. Which of the following is the correct way to declare a tree node struct in Go with integer data and pointers to left and right children?

easy

A. type Node struct { data int, left &Node, right &Node }

B. type Node struct { data int, left *Node, right *Node }

C. type Node struct { data int, left Node, right Node }

D. type Node struct { data int, left Node*, right Node* }

Solution

1. Step 1: Recall Go pointer syntax

   In Go, pointers are declared with * before the type, e.g., *Node.

2. Step 2: Check each option's pointer usage

   type Node struct { data int, left *Node, right *Node } uses *Node correctly. type Node struct { data int, left Node, right Node } uses Node (value, not pointer). type Node struct { data int, left &Node, right &Node } uses &Node which is invalid syntax. type Node struct { data int, left Node*, right Node* } uses Node* which is invalid in Go.

3. Final Answer:

   type Node struct { data int, left *Node, right *Node } -> Option B

4. Quick Check:

   Pointer syntax in Go = *Type [OK]

Hint: Use *Type for pointers in Go structs [OK]

Common Mistakes:

• Using value type instead of pointer for children
• Using &Type or Type* which are invalid in Go
• Using semicolons instead of commas in struct fields

3. Given the following tree structure in Go, what is the depth of node with value 5?

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

root := &Node{data: 1}
root.left = &Node{data: 2}
root.right = &Node{data: 3}
root.left.left = &Node{data: 4}
root.left.right = &Node{data: 5}

medium

A. 2

B. 1

C. 3

D. 4

Solution

1. Step 1: Understand depth definition

   Depth of a node is the number of edges from the root to that node.

2. Step 2: Trace path from root to node 5

   Path: 1 (root) -> 2 -> 5. Number of edges = 2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Depth = edges from root = 2 [OK]

Hint: Count edges from root to node for depth [OK]

Common Mistakes:

• Counting nodes instead of edges
• Confusing depth with height
• Miscounting path length

4. Consider this Go code snippet for a binary tree node:

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

func isLeaf(n *Node) bool {
  return n.left == nil && n.right == nil
}

root := &Node{data: 10}
root.left = &Node{data: 20}
root.right = &Node{data: 30}
root.left.left = &Node{data: 40}

fmt.Println(isLeaf(root.left))

What is the output and why?

medium

A. true

B. Runtime panic

C. Compilation error

D. false

Solution

1. Step 1: Understand isLeaf function

   It returns true if both left and right children are nil (no children).

2. Step 2: Check root.left node children

   root.left has left child (40) but no right child, so left != nil, right == nil.

3. Step 3: Evaluate isLeaf(root.left)

   Since left is not nil, isLeaf returns false.

4. Final Answer:

   false -> Option D

5. Quick Check:

   Leaf = no children, root.left has left child [OK]

Hint: Leaf nodes have no children; check both left and right nil [OK]

Common Mistakes:

• Assuming node with one child is leaf
• Confusing nil checks causing panic
• Expecting true without checking both children

5. You have a binary tree where the root is at level 0. If a node is at level 3, what is its height assuming the tree height is 5? Also, which node is at level 0? Choose the correct pair of answers.

hard

A. Height = 5; Root node is at level 0

B. Height = 3; Leaf node is at level 0

C. Height = 2; Root node is at level 0

D. Height = 2; Leaf node is at level 0

Solution

1. Step 1: Calculate height of node at level 3

   Height = total tree height - node level = 5 - 3 = 2.

2. Step 2: Identify node at level 0

   By definition, root node is at level 0.

3. Final Answer:

   Height = 2; Root node is at level 0 -> Option C

4. Quick Check:

   Height = 5 - 3 = 2; Root at level 0 [OK]

Hint: Height = tree height - level; root always level 0 [OK]

Common Mistakes:

• Confusing height with level
• Assuming leaf node at level 0
• Calculating height as level number