Lowest Common Ancestor in Binary Tree in DSA Go - 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 grows as the tree gets bigger.

How does the search through the tree affect the total work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
    if root == nil || root == p || root == q {
        return root
    }
    left := lowestCommonAncestor(root.Left, p, q)
    right := lowestCommonAncestor(root.Right, p, q)
    if left != nil && right != nil {
        return root
    }
    if left != nil {
        return left
    }
    return 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 checked once during the recursion.

How Execution Grows With Input

As the tree grows, the function visits every node once to find the answer.

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

Pattern observation: The work grows directly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the lowest common ancestor grows linearly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "The function only checks a small part of the tree, so it runs faster than O(n)."

[OK] Correct: The function may need to visit every node in the worst case to find both targets, so it must check the whole tree.

Interview Connect

Understanding this linear time complexity helps you explain why this approach is efficient and expected when working with binary trees.

Self-Check

"What if the tree was a binary search tree? 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 node with the smallest value between the two nodes

B. The root node of the binary tree

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

D. The node that is a child of both nodes

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: Identify the deepest common ancestor

   Among all common ancestors, the LCA is the one farthest from the root (deepest).

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 lowest node common to both paths [OK]

Common Mistakes:

• Confusing LCA with root node always
• Thinking LCA is the node with smaller value
• Assuming LCA is a child of both nodes

2. Which of the following is the correct Go function signature to find the LCA of two nodes in a binary tree?

easy

A. func lowestCommonAncestor(root TreeNode, p *TreeNode, q TreeNode) *TreeNode

B. func lowestCommonAncestor(root TreeNode, p TreeNode, q TreeNode) TreeNode

C. func lowestCommonAncestor(root *TreeNode, p TreeNode, q *TreeNode) TreeNode

D. func lowestCommonAncestor(root *TreeNode, p *TreeNode, q *TreeNode) *TreeNode

Solution

1. Step 1: Understand pointer usage in Go for tree nodes

   Tree nodes are usually passed as pointers to avoid copying and to allow modification.

2. Step 2: Check parameter and return types

   The function should accept pointers to TreeNode for root, p, and q, and return a pointer to TreeNode.

3. Final Answer:

   func lowestCommonAncestor(root *TreeNode, p *TreeNode, q *TreeNode) *TreeNode -> Option D

4. Quick Check:

   Use pointers for nodes in Go function signature [OK]

Hint: Use pointers for tree nodes in Go functions [OK]

Common Mistakes:

• Passing nodes by value instead of pointer
• Mixing pointer and value types in parameters
• Returning value instead of pointer

3. Given the following Go code snippet, what will be the output when finding the LCA of nodes 5 and 1?

type TreeNode struct {
    Val int
    Left, Right *TreeNode
}

func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
    if root == nil || root == p || root == q {
        return root
    }
    left := lowestCommonAncestor(root.Left, p, q)
    right := lowestCommonAncestor(root.Right, p, q)
    if left != nil && right != nil {
        return root
    }
    if left != nil {
        return left
    }
    return right
}

func main() {
    root := &TreeNode{Val: 3}
    root.Left = &TreeNode{Val: 5}
    root.Right = &TreeNode{Val: 1}
    root.Left.Left = &TreeNode{Val: 6}
    root.Left.Right = &TreeNode{Val: 2}
    root.Right.Left = &TreeNode{Val: 0}
    root.Right.Right = &TreeNode{Val: 8}

    lca := lowestCommonAncestor(root, root.Left, root.Right)
    fmt.Println(lca.Val)
}

medium

A. 3

B. 5

C. 1

D. 2

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.

2. Step 2: Both left and right recursive calls return non-nil nodes

   Since left and right are both non-nil, root (3) is the LCA.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   LCA(5,1) = 3 [OK]

Hint: If both sides return nodes, root is LCA [OK]

Common Mistakes:

• Returning one of the nodes instead of root
• Confusing node values with LCA
• Not handling nil checks properly

4. The following Go function is intended to find the LCA of two nodes in a binary tree. What is the bug?

func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
    if root == nil {
        return nil
    }
    if root.Val == p.Val || root.Val == q.Val {
        return root
    }
    left := lowestCommonAncestor(root.Left, p, q)
    right := lowestCommonAncestor(root.Right, p, q)
    if left != nil {
        return left
    }
    if right != nil {
        return right
    }
    return root
}

medium

A. Returns root when both left and right are nil instead of nil

B. Does not check for nil root at start

C. Compares node pointers instead of values

D. Does not recurse on left subtree

Solution

1. Step 1: Analyze return conditions after recursion

   If both left and right are nil, the function incorrectly returns root instead of nil.

2. Step 2: Understand correct base case for no ancestor found

   When neither subtree contains p or q, function should return nil, not root.

3. Final Answer:

   Returns root when both left and right are nil instead of nil -> Option A

4. Quick Check:

   Return nil if no ancestor found [OK]

Hint: Return nil if both sides are nil, not root [OK]

Common Mistakes:

• Returning root instead of nil when no match
• Missing nil check at start
• Comparing values instead of pointers incorrectly

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 given nodes p and q?

hard

A. Compare node values only since duplicates don't affect LCA

B. Compare node pointers instead of node values to identify nodes

C. Use node values and parent pointers to find LCA

D. Return the node with smaller value as LCA

Solution

1. Step 1: Understand problem with duplicate values

   When values are duplicated, comparing values alone can confuse nodes.

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

   Comparing pointers ensures the exact nodes p and q are found, not just matching values.

3. Step 3: Modify function to compare pointers (e.g., root == p)

   This guarantees correct LCA even with duplicate values.

4. Final Answer:

   Compare node pointers instead of node values to identify nodes -> Option B

5. Quick Check:

   Use pointers, not values, for unique node identity [OK]

Hint: Use pointers to identify nodes uniquely with duplicates [OK]

Common Mistakes:

• Comparing only values causing wrong LCA
• Assuming duplicates don't affect LCA
• Using parent pointers unnecessarily