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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of LCA function for given nodes

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

DSA Javascript

class TreeNode {
  constructor(val, left = null, right = null) {
    this.val = val;
    this.left = left;
    this.right = right;
  }
}

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

const root = new TreeNode(3);
root.left = new TreeNode(5);
root.right = new TreeNode(1);
root.left.left = new TreeNode(6);
root.left.right = new TreeNode(2);
root.right.left = new TreeNode(0);
root.right.right = new TreeNode(8);

const p = root.left; // Node with val 5
const q = root.right; // Node with val 1

const lca = lowestCommonAncestor(root, p, q);
console.log(lca.val);

A 3

B 5

C 1

D null

💡 Hint

Think about where the paths to nodes 5 and 1 meet first in the tree.

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

🧠 Conceptual

intermediate

Understanding LCA when one node is ancestor of the other

In a binary tree, if one node is the ancestor of the other, what will the Lowest Common Ancestor (LCA) function return when called with these two nodes?

A The descendant node

B The ancestor node

C The root node of the tree

D null

💡 Hint

Remember the LCA is the lowest node that has both nodes as descendants (a node can be a descendant of itself).

If one node is ancestor of the other, the ancestor node is the LCA because it contains the descendant node in its subtree.

🔧 Debug

advanced

Identify the error in this LCA implementation

What error will this code produce when trying to find the LCA of nodes 5 and 1?

DSA Javascript

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

// Usage with nodes not from the tree
const p = { val: 5 };
const q = { val: 1 };

A Infinite recursion causing stack overflow

B Throws TypeError because root.left is undefined

C Returns root node always

D Returns null because p and q are not actual tree nodes

💡 Hint

Check if the nodes p and q are the same objects as in the tree or just objects with same values.

The function compares nodes by reference (object identity), but p and q are new objects with same values, so the function never finds them and returns null.

❓ Predict Output

advanced

Output of LCA for nodes in left subtree

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

DSA Javascript

class TreeNode {
  constructor(val, left = null, right = null) {
    this.val = val;
    this.left = left;
    this.right = right;
  }
}

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

const root = new TreeNode(3);
root.left = new TreeNode(5);
root.right = new TreeNode(1);
root.left.left = new TreeNode(6);
root.left.right = new TreeNode(2);
root.right.left = new TreeNode(0);
root.right.right = new TreeNode(8);

const p = root.left.left; // Node with val 6
const q = root.left.right; // Node with val 2

const lca = lowestCommonAncestor(root, p, q);
console.log(lca.val);

A 5

B 3

C 6

D 2

💡 Hint

Both nodes are in the left subtree of root. Which node is their common ancestor?

Nodes 6 and 2 are children of node 5, so 5 is their lowest common ancestor.

🧠 Conceptual

expert

LCA in Binary Tree vs Binary Search Tree

Which statement correctly distinguishes the Lowest Common Ancestor (LCA) problem in a Binary Tree from that in a Binary Search Tree (BST)?

A LCA algorithms are identical for both Binary Tree and BST

B In Binary Tree, LCA is always the root; in BST, it is always one of the nodes

C In BST, LCA can be found using node values and tree ordering; in Binary Tree, structure must be traversed fully

D BST does not have a Lowest Common Ancestor concept

💡 Hint

Think about how BST ordering helps find LCA faster.

BST property allows comparing node values to decide direction, reducing search space. Binary Tree lacks ordering, so full traversal is needed.

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