Lowest Common Ancestor in Binary Tree in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the current node is nil.

DSA Go

if root [1] nil {
    return nil
}

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

A==

B!=

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' instead of '==' causes wrong logic.

Using '<' or '>' with pointers is invalid.

We check if the current node is nil using ==. This means no node exists here.

2fill in blank

medium

Complete the code to check if the current node matches either target node p or q.

DSA Go

if root == [1] || root == [2] {
    return root
}

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

Ap

Bq

Cnil

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'nil' instead of target nodes.

Using 'root' instead of 'p' or 'q'.

We check if the current node is equal to p or q. If yes, return it.

3fill in blank

hard

Fix the error in the recursive calls to search left and right subtrees.

DSA Go

left := lowestCommonAncestor(root.[1], p, q)
right := lowestCommonAncestor(root.[2], p, q)

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

ALeft

BRight

Cleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using uppercase 'Left' or 'Right' causes compilation errors.

Mixing field names with variable names.

In Go, struct fields are case-sensitive. The left child is left and right child is right.

4fill in blank

hard

Fill both blanks to return the correct node based on left and right subtree results.

DSA Go

if left != nil && right != nil {
    return [1]
}
if left != nil {
    return [2]
}
return right

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

Aroot

Bleft

Cright

Dnil

Attempts:

3 left

💡 Hint

Common Mistakes

Returning left or right when both are non-nil.

Returning nil incorrectly.

If both left and right are not nil, current node is the ancestor. Otherwise, return the non-nil child.

5fill in blank

hard

Fill all three blanks to complete the full function for Lowest Common Ancestor.

DSA Go

func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
    if root [1] nil {
        return nil
    }
    if root == p || root == q {
        return root
    }
    left := lowestCommonAncestor(root.[2], p, q)
    right := lowestCommonAncestor(root.[3], p, q)
    if left != nil && right != nil {
        return root
    }
    if left != nil {
        return left
    }
    return right
}

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

A==

Bleft

Cright

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' instead of '==' for nil check.

Using uppercase 'Left' or 'Right' fields.

Returning wrong nodes when both children are non-nil.

This function checks if the root is nil, then if root matches p or q. It searches left and right subtrees recursively. If both sides find targets, root is the ancestor. Otherwise, returns the non-nil child.

Practice

1. What is the Lowest Common Ancestor (LCA) of two nodes in a binary tree?

easy

A. The node with the smallest value between the two nodes

B. The root node of the binary tree

C. The deepest node that is an ancestor of both nodes

D. The node that is a child of both nodes

Solution

1. Step 1: Understand the definition of LCA

   The Lowest Common Ancestor is the lowest node in the tree that has both given nodes as descendants (a node can be a descendant of itself).

2. Step 2: Identify the deepest common ancestor

   Among all common ancestors, the LCA is the one farthest from the root (deepest).

3. Final Answer:

   The deepest node that is an ancestor of both nodes -> Option C

4. Quick Check:

   LCA = deepest common ancestor [OK]

Hint: LCA is the lowest node common to both paths [OK]

Common Mistakes:

• Confusing LCA with root node always
• Thinking LCA is the node with smaller value
• Assuming LCA is a child of both nodes

2. Which of the following is the correct Go function signature to find the LCA of two nodes in a binary tree?

easy

A. func lowestCommonAncestor(root TreeNode, p *TreeNode, q TreeNode) *TreeNode

B. func lowestCommonAncestor(root TreeNode, p TreeNode, q TreeNode) TreeNode

C. func lowestCommonAncestor(root *TreeNode, p TreeNode, q *TreeNode) TreeNode

D. func lowestCommonAncestor(root *TreeNode, p *TreeNode, q *TreeNode) *TreeNode

Solution

1. Step 1: Understand pointer usage in Go for tree nodes

   Tree nodes are usually passed as pointers to avoid copying and to allow modification.

2. Step 2: Check parameter and return types

   The function should accept pointers to TreeNode for root, p, and q, and return a pointer to TreeNode.

3. Final Answer:

   func lowestCommonAncestor(root *TreeNode, p *TreeNode, q *TreeNode) *TreeNode -> Option D

4. Quick Check:

   Use pointers for nodes in Go function signature [OK]

Hint: Use pointers for tree nodes in Go functions [OK]

Common Mistakes:

• Passing nodes by value instead of pointer
• Mixing pointer and value types in parameters
• Returning value instead of pointer

3. Given the following Go code snippet, what will be the output when finding the LCA of nodes 5 and 1?

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

func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
    if root == nil || root == p || root == q {
        return root
    }
    left := lowestCommonAncestor(root.Left, p, q)
    right := lowestCommonAncestor(root.Right, p, q)
    if left != nil && right != nil {
        return root
    }
    if left != nil {
        return left
    }
    return right
}

func main() {
    root := &TreeNode{Val: 3}
    root.Left = &TreeNode{Val: 5}
    root.Right = &TreeNode{Val: 1}
    root.Left.Left = &TreeNode{Val: 6}
    root.Left.Right = &TreeNode{Val: 2}
    root.Right.Left = &TreeNode{Val: 0}
    root.Right.Right = &TreeNode{Val: 8}

    lca := lowestCommonAncestor(root, root.Left, root.Right)
    fmt.Println(lca.Val)
}

medium

A. 3

B. 5

C. 1

D. 2

Solution

1. Step 1: Trace recursive calls for nodes 5 and 1

   Starting at root (3), left subtree contains node 5, right subtree contains node 1.

2. Step 2: Both left and right recursive calls return non-nil nodes

   Since left and right are both non-nil, root (3) is the LCA.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   LCA(5,1) = 3 [OK]

Hint: If both sides return nodes, root is LCA [OK]

Common Mistakes:

• Returning one of the nodes instead of root
• Confusing node values with LCA
• Not handling nil checks properly

4. The following Go function is intended to find the LCA of two nodes in a binary tree. What is the bug?

func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
    if root == nil {
        return nil
    }
    if root.Val == p.Val || root.Val == q.Val {
        return root
    }
    left := lowestCommonAncestor(root.Left, p, q)
    right := lowestCommonAncestor(root.Right, p, q)
    if left != nil {
        return left
    }
    if right != nil {
        return right
    }
    return root
}

medium

A. Returns root when both left and right are nil instead of nil

B. Does not check for nil root at start

C. Compares node pointers instead of values

D. Does not recurse on left subtree

Solution

1. Step 1: Analyze return conditions after recursion

   If both left and right are nil, the function incorrectly returns root instead of nil.

2. Step 2: Understand correct base case for no ancestor found

   When neither subtree contains p or q, function should return nil, not root.

3. Final Answer:

   Returns root when both left and right are nil instead of nil -> Option A

4. Quick Check:

   Return nil if no ancestor found [OK]

Hint: Return nil if both sides are nil, not root [OK]

Common Mistakes:

• Returning root instead of nil when no match
• Missing nil check at start
• Comparing values instead of pointers incorrectly

5. You have a binary tree where nodes can have duplicate values. How should you modify the LCA function to correctly find the LCA of two given nodes p and q?

hard

A. Compare node values only since duplicates don't affect LCA

B. Compare node pointers instead of node values to identify nodes

C. Use node values and parent pointers to find LCA

D. Return the node with smaller value as LCA

Solution

1. Step 1: Understand problem with duplicate values

   When values are duplicated, comparing values alone can confuse nodes.

2. Step 2: Use node pointers to uniquely identify nodes

   Comparing pointers ensures the exact nodes p and q are found, not just matching values.

3. Step 3: Modify function to compare pointers (e.g., root == p)

   This guarantees correct LCA even with duplicate values.

4. Final Answer:

   Compare node pointers instead of node values to identify nodes -> Option B

5. Quick Check:

   Use pointers, not values, for unique node identity [OK]

Hint: Use pointers to identify nodes uniquely with duplicates [OK]

Common Mistakes:

• Comparing only values causing wrong LCA
• Assuming duplicates don't affect LCA
• Using parent pointers unnecessarily