Lowest Common Ancestor in Binary Tree in DSA Python - Time & Space Complexity

Time Complexity: Lowest Common Ancestor in Binary Tree

O(n)

Understanding Time Complexity

We want to understand how the time needed to find the lowest common ancestor changes as the tree grows.

How does the search through the tree affect the time taken?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code finds the lowest common ancestor of two nodes in a binary tree by searching both sides recursively.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls visiting each node once.
• How many times: Each node is visited once during the search.

How Execution Grows With Input

As the tree grows, the function visits more nodes, roughly once per node.

Input Size (n)	Approx. Operations
10	About 10 visits
100	About 100 visits
1000	About 1000 visits

Pattern observation: The operations grow linearly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the ancestor grows directly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "The search only looks at part of the tree, so it is faster than visiting all nodes."

[OK] Correct: The function may need to check every node in the worst case, especially if the nodes are deep or on opposite sides.

Interview Connect

Understanding this helps you explain how recursive tree searches work and why they take time proportional to tree size.

Self-Check

"What if the tree was a binary search tree and you used its properties? How would the time complexity change?"

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