Lowest Common Ancestor in Binary Tree in DSA C# - Cheat Sheet & Quick Revision

Recall & Review

beginner

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

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

beginner

In a binary tree, if one node is ancestor of the other, what is the LCA?

The LCA is the ancestor node itself because a node can be a descendant of itself.

intermediate

What is the main idea behind the recursive approach to find LCA in a binary tree?

Recursively search left and right subtrees for the two nodes. If both sides return non-null, current node is LCA. If only one side returns non-null, propagate that up.

intermediate

What is the time complexity of finding LCA in a binary tree using the recursive approach?

O(n), where n is the number of nodes in the tree, because each node is visited once in the worst case.

advanced

Why does the LCA algorithm work even if the two nodes are not guaranteed to be in the tree?

Because the recursive calls return null if nodes are not found, and the algorithm only returns a node if both nodes are found in its subtree, otherwise null is returned.

What does the Lowest Common Ancestor represent in a binary tree?

A The leaf node closest to the root

B The root node of the tree

C The node with the smallest value

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

The LCA is the deepest node that has both nodes as descendants.

If one node is the ancestor of the other, what is the LCA?

A The ancestor node itself

B The root node

C The other node

D There is no LCA

A node can be a descendant of itself, so the ancestor node is the LCA.

What is the base case in the recursive LCA algorithm?

A When the current node is null or matches one of the target nodes

B When both children are null

C When the current node is the root

D When the tree is empty

The recursion stops when it finds a null node or one of the target nodes.

What is the time complexity of the recursive LCA algorithm in a binary tree?

A O(log n)

B O(n log n)

C O(n)

D O(1)

Each node is visited once, so the complexity is linear in the number of nodes.

If neither node is found in the tree, what does the LCA function return?

A One of the nodes

B Null

C The root node

D An error

The function returns null if neither node is found.

Explain how the recursive approach finds the Lowest Common Ancestor in a binary tree.

Think about how recursion explores both sides and combines results.

Describe the edge cases to consider when implementing the Lowest Common Ancestor algorithm in a binary tree.

Consider what happens when nodes are missing or equal.

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