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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Javascript

if (root [1] null) return null;

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

A>

B!==

C<

D===

Attempts:

3 left

💡 Hint

Common Mistakes

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

Using '>' or '<' to compare with null is incorrect.

We check if root is exactly equal to null to stop recursion when we reach a leaf's child.

2fill in blank

medium

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

DSA Javascript

if (root.val [1] p.val || root.val [1] q.val) return root;

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

A!==

B===

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

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

Using loose equality '==' can cause bugs.

We use strict equality '===' to check if the current node's value matches p or q.

3fill in blank

hard

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

DSA Javascript

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

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

Aleft

Bright

CleftChild

DrightChild

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect property names like leftChild or rightChild.

Recursing on the same subtree twice.

We must recurse on root.left and root.right to search both subtrees.

4fill in blank

hard

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

DSA Javascript

if (left [1] null && right [2] null) return root;

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 condition.

Using logical OR '||' instead of AND '&&' changes logic.

If both left and right are not null, root is the lowest common ancestor.

5fill in blank

hard

Fill the blanks to return the non-null subtree result or null.

DSA Javascript

return left [1] right ? left [2] right : null;

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

A!==

B||

C&&

D===

Attempts:

3 left

💡 Hint

Common Mistakes

Using '&&' instead of '||' changes the logic.

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

Return left if left is not null; otherwise return right if right is not null; else null.

Practice

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

easy

A. The node that appears first in an inorder traversal

B. The root node of the binary tree

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

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

Solution

1. Step 1: Understand ancestor concept

   An ancestor of a node is any node on the path from the root to that node.

2. Step 2: Define lowest common ancestor

   The lowest common ancestor is the deepest node that is an ancestor of both given nodes.

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 deepest shared parent node [OK]

Common Mistakes:

• Confusing LCA with root node
• Choosing node with smallest value
• Selecting node based on traversal order

2. Which of the following is the correct base case condition in a recursive function to find LCA in a binary tree?

easy

A. if (root.right == null) return root;

B. if (root.left == null) return root;

C. if (root.value == null) return root;

D. if (root == null) return null;

Solution

1. Step 1: Identify base case for recursion

   When the current node is null, it means we've reached the end of a path without finding the target nodes.

2. Step 2: Correct base case condition

   Return null to indicate no ancestor found in this path.

3. Final Answer:

   if (root == null) return null; -> Option D

4. Quick Check:

   Base case returns null on null node [OK]

Hint: Return null when node is null [OK]

Common Mistakes:

• Returning root when left or right child is null
• Checking root.value == null which is invalid
• Missing base case leading to infinite recursion

3. Consider the following binary tree and code snippet to find LCA of nodes 5 and 1:

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

function findLCA(root, p, q) {
  if (!root || root === p || root === q) return root;
  const left = findLCA(root.left, p, q);
  const right = findLCA(root.right, p, q);
  if (left && right) return root;
  return left ? left : right;
}

const lca = findLCA(root, node5, node1);
console.log(lca.value);

What will be printed?

medium

A. 5

B. 3

C. 1

D. null

Solution

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

   Both nodes 5 and 1 are children of root 3, so recursion finds them in left and right subtrees respectively.

2. Step 2: Determine LCA based on recursion

   Since left and right recursive calls return non-null (nodes 5 and 1), the current root 3 is the LCA.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   LCA of 5 and 1 is 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
• Ignoring both left and right recursive results

4. The following code attempts to find the LCA but has a bug. What is the error?

function findLCA(root, p, q) {
  if (!root) return null;
  if (root === p || root === q) return root;
  const left = findLCA(root.left, p, q);
  const right = findLCA(root.right, p, q);
  if (left) return left;
  if (right) return right;
  return null;
}

medium

A. It returns left child even if right child is also non-null

B. It does not check if root is null

C. It returns root instead of null when no nodes found

D. It does not handle the case when p or q is missing

Solution

1. Step 1: Analyze return conditions after recursion

   The code returns left if left is non-null without checking right, missing the case when both are non-null.

2. Step 2: Identify missing combined check

   It should return root if both left and right are non-null, indicating LCA found at current root.

3. Final Answer:

   It returns left child even if right child is also non-null -> Option A

4. Quick Check:

   Must check both sides before returning [OK]

Hint: Check both left and right before returning root [OK]

Common Mistakes:

• Returning left without checking right
• Missing combined condition for LCA
• Ignoring null root check

5. Given a binary tree where nodes 2 and 4 are in the left subtree of node 3, and nodes 5 and 6 are in the right subtree of node 3, what is the LCA of nodes 2 and 6?

hard

A. 3

B. 2

C. 4

D. 5

Solution

1. Step 1: Understand tree structure

   Nodes 2 and 4 are in left subtree of 3; nodes 5 and 6 are in right subtree of 3.

2. Step 2: Find common ancestor of nodes 2 and 6

   Since 2 is in left subtree and 6 is in right subtree, their lowest common ancestor is the root node 3.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   LCA of nodes in opposite subtrees is root [OK]

Hint: LCA of nodes in different subtrees is their root [OK]

Common Mistakes:

• Choosing one of the nodes instead of ancestor
• Confusing sibling nodes as LCA
• Ignoring tree structure