Lowest Common Ancestor in Binary Tree in DSA Typescript - 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 following TypeScript code that finds the Lowest Common Ancestor (LCA) of nodes 5 and 1 in the binary tree?

DSA Typescript

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
    this.val = val === undefined ? 0 : val;
    this.left = left === undefined ? null : left;
    this.right = right === undefined ? null : right;
  }
}

function lowestCommonAncestor(root: TreeNode | null, p: TreeNode, q: TreeNode): TreeNode | null {
  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;
}

// Construct the tree
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);
root.left.right.left = new TreeNode(7);
root.left.right.right = new TreeNode(4);

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

const ancestor = lowestCommonAncestor(root, p, q);
console.log(ancestor ? ancestor.val : null);

A null

B 5

C 1

D 3

💡 Hint

Think about where nodes 5 and 1 meet first when traversing from the root.

The LCA of nodes 5 and 1 is the root node 3 because 5 is in the left subtree and 1 is in the right subtree. The function returns the first node where both nodes are found in different subtrees.

❓ Predict Output

intermediate

Output of LCA for nodes in same subtree

What is the output of the following TypeScript code that finds the Lowest Common Ancestor (LCA) of nodes 7 and 4 in the binary tree?

DSA Typescript

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
    this.val = val === undefined ? 0 : val;
    this.left = left === undefined ? null : left;
    this.right = right === undefined ? null : right;
  }
}

function lowestCommonAncestor(root: TreeNode | null, p: TreeNode, q: TreeNode): TreeNode | null {
  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;
}

// Construct the tree
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);
root.left.right.left = new TreeNode(7);
root.left.right.right = new TreeNode(4);

const p = root.left.right.left; // Node with val 7
const q = root.left.right.right; // Node with val 4

const ancestor = lowestCommonAncestor(root, p, q);
console.log(ancestor ? ancestor.val : null);

A 5

B 2

C 3

D 7

💡 Hint

Both nodes are in the right subtree of node 5.

Nodes 7 and 4 are both children of node 2, so their lowest common ancestor is node 2.

🔧 Debug

advanced

Identify the error in LCA function

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

DSA Typescript

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
    this.val = val === undefined ? 0 : val;
    this.left = left === undefined ? null : left;
    this.right = right === undefined ? null : right;
  }
}

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

// Construct the tree
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);
root.left.right.left = new TreeNode(7);
root.left.right.right = new TreeNode(4);

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

const ancestor = lowestCommonAncestor(root, p, q);
console.log(ancestor ? ancestor.val : null);

A No error, outputs 3

B TypeError: Cannot read property 'val' of null

C Outputs null

D Stack overflow due to infinite recursion

💡 Hint

Check how the function compares nodes using val property.

The function correctly checks for null and compares node values. It returns the root 3 as the LCA without error.

❓ Predict Output

advanced

Output when one node is ancestor of the other

What is the output of the following TypeScript code that finds the Lowest Common Ancestor (LCA) of nodes 5 and 4, where 5 is an ancestor of 4?

DSA Typescript

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
    this.val = val === undefined ? 0 : val;
    this.left = left === undefined ? null : left;
    this.right = right === undefined ? null : right;
  }
}

function lowestCommonAncestor(root: TreeNode | null, p: TreeNode, q: TreeNode): TreeNode | null {
  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;
}

// Construct the tree
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);
root.left.right.left = new TreeNode(7);
root.left.right.right = new TreeNode(4);

const p = root.left; // Node with val 5
const q = root.left.right.right; // Node with val 4

const ancestor = lowestCommonAncestor(root, p, q);
console.log(ancestor ? ancestor.val : null);

A 3

B 4

C 5

D null

💡 Hint

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

Since node 5 is an ancestor of node 4, the LCA is node 5 itself.

🧠 Conceptual

expert

Understanding LCA in a binary tree with no parent pointers

In a binary tree where nodes do not have parent pointers, which of the following statements about the Lowest Common Ancestor (LCA) algorithm is TRUE?

A The LCA algorithm must use recursion or a stack to explore subtrees from the root downwards.

B The LCA can be found by sorting the node values and picking the middle value.

C The LCA is always the root node of the tree regardless of input nodes.

D The LCA can be found by traversing from each node up to the root and comparing paths.

💡 Hint

Without parent pointers, you cannot move upwards from a node easily.

Without parent pointers, the algorithm must start from the root and explore subtrees recursively or with a stack to find the LCA.

Practice

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

easy

A. The root node of the binary tree

B. The lowest node that has both given nodes as descendants

C. The node with the highest value in the tree

D. The leaf node closest to the root

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: Differentiate from other nodes

   It is not necessarily the root or the highest value node, but the shared ancestor closest to the two nodes.

3. Final Answer:

   The lowest node that has both given nodes as descendants -> Option B

4. Quick Check:

   LCA = lowest shared ancestor [OK]

Hint: LCA is the closest shared parent node [OK]

Common Mistakes:

• Confusing LCA with root node
• Thinking LCA is the node with highest value
• Assuming LCA must be a leaf node

2. Which of the following is the correct TypeScript function signature to find the LCA in a binary tree node structure?

easy

A. function findLCA(root: TreeNode, p: TreeNode, q: TreeNode): TreeNode | null

B. function findLCA(root: TreeNode, p: number, q: number): number

C. function findLCA(root: TreeNode): TreeNode

D. function findLCA(p: TreeNode, q: TreeNode): TreeNode

Solution

1. Step 1: Analyze parameters needed for LCA

   To find LCA, we need the root of the tree and the two nodes (p and q) to find the ancestor for.

2. Step 2: Check return type

   The function should return the TreeNode that is the LCA or null if not found.

3. Final Answer:

   function findLCA(root: TreeNode, p: TreeNode, q: TreeNode): TreeNode | null -> Option A

4. Quick Check:

   Correct signature includes root, p, q nodes and returns TreeNode or null [OK]

Hint: LCA function needs root and two nodes as input [OK]

Common Mistakes:

• Missing root parameter
• Using wrong parameter types like numbers instead of nodes
• Returning wrong type like number instead of TreeNode

3. Given the following TypeScript code for LCA, what will be the output if we call findLCA(root, node5, node1) where node5 and node1 are nodes with values 5 and 1 respectively?

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val: number) {
    this.val = val;
    this.left = null;
    this.right = null;
  }
}

function findLCA(root: TreeNode | null, p: TreeNode, q: TreeNode): TreeNode | null {
  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;
}

// Tree structure:
//       3
//      / \
//     5   1
//    / \ / \
//   6  2 0  8
//     / \
//    7   4

const root = new TreeNode(3);
const node5 = new TreeNode(5);
const node1 = new TreeNode(1);
root.left = node5;
root.right = node1;
// ... other nodes assigned similarly

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

medium

A. 3

B. 5

C. 1

D. null

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. Both left and right recursive calls return non-null nodes.

2. Step 2: Determine LCA based on recursive returns

   Since both left and right calls return nodes, current root (3) is the LCA.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   LCA of 5 and 1 is 3 [OK]

Hint: If nodes are in different subtrees, LCA is root [OK]

Common Mistakes:

• Returning one of the nodes instead of their ancestor
• Confusing node values with node references
• Ignoring null checks in recursion

4. Identify the bug in the following TypeScript LCA function and select the correct fix:

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

medium

A. Add a base case to return root when root is null

B. Replace logical OR with AND in return statement

C. Change comparisons to check node references, not values

D. Remove recursive calls to left and right subtrees

Solution

1. Step 1: Analyze node comparison logic

   The function compares node values (root.val === p.val), but nodes should be compared by reference to handle duplicates correctly.

2. Step 2: Correct the comparison to use reference equality

   Change to (root === p || root === q) to compare actual nodes, not just values.

3. Final Answer:

   Change comparisons to check node references, not values -> Option C

4. Quick Check:

   Node reference comparison fixes LCA correctness [OK]

Hint: Compare nodes by reference, not by value [OK]

Common Mistakes:

• Comparing node values instead of references
• Misusing logical operators in return
• Removing necessary recursion

5. You have a binary tree where nodes can have duplicate values. Which approach ensures correct Lowest Common Ancestor (LCA) identification for two given nodes p and q?

hard

A. Compare nodes by their values during recursion

B. Ignore duplicates and return the root as LCA

C. Return the node with the smaller value as LCA

D. Use node references (memory addresses) for comparison

Solution

1. Step 1: Understand the problem with duplicate values

   When values repeat, comparing nodes by value can lead to wrong LCA because multiple nodes share the same value.

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

   Comparing nodes by their references (actual objects) ensures the correct nodes are identified regardless of value duplicates.

3. Final Answer:

   Use node references (memory addresses) for comparison -> Option D

4. Quick Check:

   Reference comparison handles duplicates correctly [OK]

Hint: Use node references to handle duplicates safely [OK]

Common Mistakes:

• Comparing only node values
• Assuming smaller value node is LCA
• Ignoring duplicates in tree