Lowest Common Ancestor in Binary Tree in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of LCA for given nodes in a binary tree

What is the output of the program when finding the Lowest Common Ancestor (LCA) of nodes 5 and 1 in the given binary tree?

DSA Go

package main

import "fmt"

type TreeNode struct {
    Val   int
    Left  *TreeNode
    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}

    p := root.Left      // Node with Val 5
    q := root.Right     // Node with Val 1

    lca := lowestCommonAncestor(root, p, q)
    fmt.Println(lca.Val)
}

A 3

B 5

C 1

D 6

💡 Hint

Think about the lowest node that has both nodes 5 and 1 as descendants.

The Lowest Common Ancestor of nodes 5 and 1 is the root node 3 because 3 is the lowest node in the tree that has both 5 and 1 as descendants.

❓ Predict Output

intermediate

Output of LCA for nodes in the same subtree

What is the output of the program when finding the Lowest Common Ancestor (LCA) of nodes 6 and 2 in the given binary tree?

DSA Go

package main

import "fmt"

type TreeNode struct {
    Val   int
    Left  *TreeNode
    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}

    p := root.Left.Left   // Node with Val 6
    q := root.Left.Right  // Node with Val 2

    lca := lowestCommonAncestor(root, p, q)
    fmt.Println(lca.Val)
}

A 2

B 6

C 5

D 3

💡 Hint

Both nodes are in the left subtree of the root.

Nodes 6 and 2 are both children of node 5, so the LCA is node 5.

🔧 Debug

advanced

Identify the error in LCA function implementation

What error will occur when running this LCA function code?

DSA Go

func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
    if root == nil {
        return nil
    }
    if 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
}

A Panic due to nil pointer dereference if p or q is nil

B No error, works correctly

C Infinite recursion causing stack overflow

D Compilation error due to missing return statement

💡 Hint

Check how the function handles nil and base cases.

The function correctly handles nil root and base cases by returning root when it matches p or q. It will not cause runtime or compile errors.

❓ Predict Output

advanced

Output when one node is ancestor of the other

What is the output of the program when finding the Lowest Common Ancestor (LCA) of nodes 5 and 6 in the given binary tree?

DSA Go

package main

import "fmt"

type TreeNode struct {
    Val   int
    Left  *TreeNode
    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}

    p := root.Left      // Node with Val 5
    q := root.Left.Left // Node with Val 6

    lca := lowestCommonAncestor(root, p, q)
    fmt.Println(lca.Val)
}

A 6

B 3

C 2

D 5

💡 Hint

If one node is ancestor of the other, the ancestor is the LCA.

Node 5 is an ancestor of node 6, so the LCA is node 5.

🧠 Conceptual

expert

Understanding LCA algorithm behavior with missing nodes

If one of the nodes (p or q) is not present in the binary tree, what will the LCA function return?

A Returns the node that is present if found, otherwise nil

B Returns nil because both nodes must be present to find LCA

C Returns the root node always regardless of presence

D Causes a runtime error due to nil pointer dereference

💡 Hint

Consider how the function returns when it finds p or q and when it doesn't.

The function returns the node if found; if only one node is present, it returns that node as the LCA. If neither is found, it returns nil.

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