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

DSA Python

class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

def lca(root, p, q):
    if root is None or root.val == p or root.val == q:
        return root
    left = lca(root.left, p, q)
    right = lca(root.right, p, q)
    if left and right:
        return root
    return left if left else right

# Construct the tree
root = Node(3)
root.left = Node(5)
root.right = Node(1)
root.left.left = Node(6)
root.left.right = Node(2)
root.right.left = Node(0)
root.right.right = Node(8)
root.left.right.left = Node(7)
root.left.right.right = Node(4)

ancestor = lca(root, 5, 1)
print(ancestor.val if ancestor else None)

A None

B 5

C 1

D 3

💡 Hint

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

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 of 3.

❓ Predict Output

intermediate

Output of LCA for nodes in same subtree

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

DSA Python

class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

def lca(root, p, q):
    if root is None or root.val == p or root.val == q:
        return root
    left = lca(root.left, p, q)
    right = lca(root.right, p, q)
    if left and right:
        return root
    return left if left else right

# Construct the tree
root = Node(3)
root.left = Node(5)
root.right = Node(1)
root.left.left = Node(6)
root.left.right = Node(2)
root.right.left = Node(0)
root.right.right = Node(8)
root.left.right.left = Node(7)
root.left.right.right = Node(4)

ancestor = lca(root, 7, 4)
print(ancestor.val if ancestor else None)

A 4

B 5

C 2

D 7

💡 Hint

Both nodes are in the right subtree of node 5.

Nodes 7 and 4 are both children of node 2, so the LCA is node 2.

🔧 Debug

advanced

Identify the error in LCA function

What error will the following code raise when trying to find the LCA of nodes 5 and 6?

DSA Python

class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

def lca(root, p, q):
    if root is None:
        return None
    if root.val == p:
        return root
    if root.val == q:
        return root
    left = lca(root.left, p, q)
    right = lca(root.right, p, q)
    if left and right:
        return root
    return left if left else right

root = Node(3)
root.left = Node(5)
root.right = Node(1)
ancestor = lca(root, 5, 6)
print(ancestor.val)

A TypeError: unsupported operand type(s) for +: 'int' and 'NoneType'

B AttributeError: 'int' object has no attribute 'val'

C None

D 3

💡 Hint

Check what the function returns when root.val matches p or q.

The function returns the integer p or q instead of the node object, so accessing ancestor.val causes AttributeError.

🧠 Conceptual

advanced

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?

A The ancestor node itself

B The descendant node

C The root node of the tree

D None

💡 Hint

Think about the definition of ancestor and how the LCA function works.

If one node is ancestor of the other, the LCA is the ancestor node itself because it is the lowest node that contains both.

🚀 Application

expert

Find LCA in a binary tree with missing nodes

Given the binary tree below, what is the output of finding the LCA of nodes 7 and 10? Tree structure: - 3 - 5 - 6 - 2 - 7 - 4 - 1 - 0 - 8 Note: Node 10 does not exist in the tree.

DSA Python

class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

def lca(root, p, q):
    if root is None or root.val == p or root.val == q:
        return root
    left = lca(root.left, p, q)
    right = lca(root.right, p, q)
    if left and right:
        return root
    return left if left else right

root = Node(3)
root.left = Node(5)
root.right = Node(1)
root.left.left = Node(6)
root.left.right = Node(2)
root.right.left = Node(0)
root.right.right = Node(8)
root.left.right.left = Node(7)
root.left.right.right = Node(4)

ancestor = lca(root, 7, 10)
print(ancestor.val if ancestor else None)

A 7

B None

C 2

D 3

💡 Hint

If one node is missing, the function returns the node found if any.

Since node 10 is missing, the function returns node 7 because it found node 7 but not 10.

Practice

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

easy

A. The root node of the binary tree

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

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

D. The parent node of the first node

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.

2. Step 2: Compare options with definition

   The deepest node that is an ancestor of both nodes matches the definition exactly. Other options describe different 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
• Thinking LCA is the smaller value node
• Assuming LCA is always the parent of one node

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

easy

A. def find_lca(root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode:

B. def find_lca(root, p, q): return int

C. def find_lca(root: int, p: int, q: int) -> TreeNode:

D. def find_lca(root: TreeNode, p: TreeNode, q: TreeNode): return str

Solution

1. Step 1: Check parameter types and return type

   The function should accept three TreeNode objects and return a TreeNode object representing the LCA.

2. Step 2: Match with options

   def find_lca(root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode: correctly uses type hints for parameters and return type. Others have wrong types or return types.

3. Final Answer:

   def find_lca(root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode: -> Option A

4. Quick Check:

   Correct signature uses TreeNode and returns TreeNode [OK]

Hint: Use TreeNode type hints for nodes and return [OK]

Common Mistakes:

• Using int instead of TreeNode for parameters
• Returning wrong type like int or str
• Omitting return type annotation

3. Given the following binary tree and code, what will be the output?

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def lowestCommonAncestor(root, p, q):
    if not root or root == p or root == q:
        return root
    left = lowestCommonAncestor(root.left, p, q)
    right = lowestCommonAncestor(root.right, p, q)
    if left and right:
        return root
    return left if left else right

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

root = TreeNode(3)
root.left = TreeNode(5)
root.right = TreeNode(1)
root.left.left = TreeNode(6)
root.left.right = TreeNode(2)
root.right.left = TreeNode(0)
root.right.right = TreeNode(8)
root.left.right.left = TreeNode(7)
root.left.right.right = TreeNode(4)

p = root.left  # Node 5
q = root.left.right.right  # Node 4

ancestor = lowestCommonAncestor(root, p, q)
print(ancestor.val)

medium

A. 4

B. 3

C. 2

D. 5

Solution

1. Step 1: Trace recursive calls for nodes 5 and 4

   Node 5 is root.left, node 4 is root.left.right.right. The LCA is the lowest node that has both as descendants. Node 5 is an ancestor of node 4.

2. Step 2: Identify the LCA node

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

3. Final Answer:

   5 -> Option D

4. Quick Check:

   LCA(5,4) = 5 [OK]

Hint: If one node is ancestor of other, LCA is that node [OK]

Common Mistakes:

• Choosing root (3) instead of ancestor node
• Confusing node values with LCA
• Ignoring that node can be ancestor of itself

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

def lowestCommonAncestor(root, p, q):
    if not root:
        return None
    if root == p or root == q:
        return root
    left = lowestCommonAncestor(root.left, p, q)
    right = lowestCommonAncestor(root.right, p, q)
    if left and right:
        return root
    if left:
        return left
    else:
        return right

# Bug: The function sometimes returns None even if both nodes exist in tree.

medium

A. The function should not return root when left and right are both found

B. The function does not check if both nodes exist in the tree before returning

C. The function should return left and right together instead of root when both exist

D. The base case should return root instead of None when root is None

Solution

1. Step 1: Analyze base cases and recursive returns

   The function returns None if root is None, and returns root if root matches p or q. It then searches left and right subtrees.

2. Step 2: Identify missing check for node existence

   If one node is not in the tree, the function may return None incorrectly. It does not verify both nodes exist before returning LCA.

3. Final Answer:

   The function does not check if both nodes exist in the tree before returning -> Option B

4. Quick Check:

   Missing node existence check causes None returns [OK]

Hint: Verify both nodes exist before returning LCA [OK]

Common Mistakes:

• Assuming both nodes always exist in tree
• Changing base case to return root when root is None
• Returning root incorrectly when both left and right found

5. You have a binary tree where nodes can have duplicate values. How should you modify the LCA function to correctly find the LCA of two nodes p and q given only their values (not references)?

hard

A. Modify the function to find all nodes with given values and find LCA of all pairs

B. Use node references instead of values to avoid ambiguity

C. Modify the function to search nodes by value and return the first matching node found

D. Return the root node always since duplicates confuse the search

Solution

1. Step 1: Understand problem with duplicate values

   When values are duplicated, searching by value alone can find multiple nodes, so LCA must consider all matching nodes.

2. Step 2: Modify approach to handle duplicates

   Find all nodes matching p and q values, then compute LCA for each pair to find the correct lowest common ancestor.

3. Final Answer:

   Modify the function to find all nodes with given values and find LCA of all pairs -> Option A

4. Quick Check:

   Handle duplicates by checking all matching nodes [OK]

Hint: Check all nodes with values to handle duplicates [OK]

Common Mistakes:

• Using only first found node causing wrong LCA
• Assuming node references are always available
• Returning root node incorrectly for duplicates