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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the current node is None.

DSA Python

if root [1] None:
    return None

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 logic.

We check if the current node is exactly None to stop recursion.

2fill in blank

medium

Complete the code to check if the current node is either p or q.

DSA Python

if root == [1] or root == q:
    return root

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

Aroot

BNone

Cp

Dq

Attempts:

3 left

💡 Hint

Common Mistakes

Using None or q in place of p in the first comparison.

We check if root matches node p or q to return it as a potential ancestor.

3fill in blank

hard

Fix the error in the recursive call to search the left subtree.

DSA Python

left = [1](root.left, p, q)

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

AlowestCommonAncestor

BlowestAncestor

CcommonAncestor

DfindAncestor

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect or partial function names causes NameError.

The function calls itself recursively by its own name to search the left subtree.

4fill in blank

hard

Fill both blanks to return the ancestor found in left or right subtree.

DSA Python

if left is not None and right is not None:
    return [1]

return [2] if left is not None else right

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

Aroot

Bleft

Cright

DNone

Attempts:

3 left

💡 Hint

Common Mistakes

Returning wrong variables causes incorrect ancestor detection.

If both left and right are found, root is the lowest common ancestor. Otherwise, return whichever subtree found a node.

5fill in blank

hard

Fill all three blanks to complete the recursive function for lowest common ancestor.

DSA Python

def lowestCommonAncestor(root, p, q):
    if root [1] None:
        return None
    if root == p or root == q:
        return root
    left = lowestCommonAncestor(root.left, p, q)
    right = lowestCommonAncestor(root.right, p, q)
    if left is not None and right is not None:
        return [2]
    return [3] if left is not None else right

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

A==

Broot

Cleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' instead of '==' in comparison.

Returning wrong variables in the final steps.

The function checks if root is None, returns root if it matches p or q, recursively searches left and right, returns root if both sides found nodes, else returns the found side.

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