Lowest Common Ancestor in Binary Tree in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA C#

if (root == [1]) return null;

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

Anull

B0

Cfalse

Dtrue

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 instead of null.

Using true or false which are boolean values.

We check if the current node is null to stop recursion.

2fill in blank

medium

Complete the code to check if the current node matches either of the target nodes.

DSA C#

if (root == p || root == [1]) return root;

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

Anull

Br

Cp

Dq

Attempts:

3 left

💡 Hint

Common Mistakes

Checking against p twice.

Checking against null or unrelated variables.

We return the current node if it matches either p or q.

3fill in blank

hard

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

DSA C#

var left = LowestCommonAncestor(root.[1], p, q);
var right = LowestCommonAncestor(root.[2], p, q);

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

AleftChild

BrightChild

Cleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect property names like leftChild or rightChild.

Using uppercase letters which are not standard.

The standard property names for child nodes are 'left' and 'right'.

4fill in blank

hard

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

DSA C#

if (left != null && right != null) return [1];
return [2] != null ? [2] : left;

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

Aroot

Bleft

Cright

Dp

Attempts:

3 left

💡 Hint

Common Mistakes

Returning left or right when both are non-null.

Returning null when one side is non-null.

If both sides are non-null, root is the ancestor; otherwise return the non-null side.

5fill in blank

hard

Fill all three blanks to complete the method signature and return type.

DSA C#

public [1] LowestCommonAncestor([2] root, [3] p, [3] q) {
    // method body
}

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

ATreeNode

Bvoid

Dint

Attempts:

3 left

💡 Hint

Common Mistakes

Using void as return type.

Using int instead of TreeNode for parameters.

The method returns a TreeNode and takes TreeNode parameters for root, p, and q.

Practice

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

easy

A. The node that is a child of both nodes

B. The root node of the binary tree

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

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

Solution

1. Step 1: Understand the definition of LCA

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

2. Step 2: Identify the correct description

   Among the options, only the deepest common ancestor fits this definition.

3. Final Answer:

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

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
• Thinking LCA is the smaller value node
• Assuming LCA is a child of both nodes

2. Which of the following is the correct C# method signature to find the LCA in a binary tree node class named TreeNode?

easy

A. public TreeNode FindLCA(TreeNode root, TreeNode p, TreeNode q)

B. public void FindLCA(TreeNode root, int p, int q)

C. private int FindLCA(int root, int p, int q)

D. public TreeNode FindLCA(int root, TreeNode p, TreeNode q)

Solution

1. Step 1: Analyze method parameters and return type

   The method should accept the root node and two nodes to find LCA for, and return a TreeNode representing the LCA.

2. Step 2: Match signature with correct types

   public TreeNode FindLCA(TreeNode root, TreeNode p, TreeNode q) correctly uses TreeNode for all nodes and returns TreeNode. Other options have incorrect parameter types or return types.

3. Final Answer:

   public TreeNode FindLCA(TreeNode root, TreeNode p, TreeNode q) -> Option A

4. Quick Check:

   Correct signature uses TreeNode parameters and return type [OK]

Hint: LCA method returns TreeNode and takes TreeNode parameters [OK]

Common Mistakes:

• Using int instead of TreeNode for nodes
• Wrong return type (void or int)
• Incorrect access modifier or parameter order

3. Given the following binary tree and code snippet, what will be the output of FindLCA(root, node5, node1).val?

    3
   / \
  5   1
 / \ / \
6  2 0  8
   / \
  7   4

// Code snippet:
TreeNode lca = FindLCA(root, node5, node1);
Console.WriteLine(lca.val);

medium

A. 5

B. 3

C. 1

D. 2

Solution

1. Step 1: Locate nodes 5 and 1 in the tree

   Node 5 is left child of root 3, node 1 is right child of root 3.

2. Step 2: Find lowest common ancestor

   The root node 3 is the lowest node that has both 5 and 1 as descendants.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   LCA(5,1) = 3 [OK]

Hint: LCA of nodes on different sides is their root ancestor [OK]

Common Mistakes:

• Choosing one of the nodes instead of their ancestor
• Confusing node values with LCA
• Ignoring tree structure

4. Identify the bug in this C# method to find LCA in a binary tree:

public TreeNode FindLCA(TreeNode root, TreeNode p, TreeNode q) {
    if (root == null) return null;
    if (root == p || root == q) return root;
    TreeNode left = FindLCA(root.left, p, q);
    TreeNode right = FindLCA(root.right, p, q);
    if (left != null && right != null) return root;
    return left != null ? left : right;
}

medium

A. Missing base case for leaf nodes

B. Should check if root.val equals p.val or q.val instead of node references

C. No bug, method is correct

D. Return statement should always return root

Solution

1. Step 1: Review base cases and recursive calls

   The method returns null if root is null, returns root if root matches p or q, then recursively searches left and right subtrees.

2. Step 2: Check logic for combining results

   If both left and right are non-null, root is LCA; else returns whichever subtree found a node. This is correct logic.

3. Final Answer:

   No bug, method is correct -> Option C

4. Quick Check:

   Standard LCA recursive method is correct [OK]

Hint: Check if both sides return non-null to find LCA [OK]

Common Mistakes:

• Comparing node values instead of references if nodes are unique
• Missing base case for null root
• Incorrect return logic

5. You have a binary tree where nodes may not be unique. How would you modify the LCA method to find the LCA of two nodes given only their values (int pVal, int qVal), assuming multiple nodes can have the same value?

hard

A. Use a method that finds paths from root to each node value and then compare paths to find LCA

B. Modify the method to return the first node found with matching value and use parent pointers to find LCA

C. Return the root node always since values are not unique

D. Use a hash set to store visited nodes and return the first common node visited

Solution

1. Step 1: Understand the problem with non-unique values

   Since values are not unique, we cannot rely on node references or values alone to find LCA directly.

2. Step 2: Use path comparison approach

   Find paths from root to each node value (may find multiple nodes), then compare paths to find the last common node, which is the LCA.

3. Final Answer:

   Use a method that finds paths from root to each node value and then compare paths to find LCA -> Option A

4. Quick Check:

   Path comparison works for non-unique values [OK]

Hint: Find paths for both values, then compare to find LCA [OK]

Common Mistakes:

• Assuming node references are unique
• Returning root always
• Using hash set without paths