Lowest Common Ancestor in Binary Tree in DSA Go

Mental Model

Find the closest shared parent node of two given nodes in a tree by checking where their paths meet.

Analogy: Imagine a family tree where you want to find the closest common grandparent of two cousins by tracing their parents upwards until they meet.

      3
     / \
    5   1
   / \ / \
  6  2 0  8
    / \
   7   4

Nodes: 5 and 1
Goal: Find their lowest common ancestor

Dry Run Walkthrough

Input: Binary tree with nodes 3,5,1,6,2,0,8,7,4; find LCA of nodes 5 and 1

Goal: Find the lowest common ancestor node of 5 and 1 in the tree

Step 1: Start at root node 3, check if it matches 5 or 1

3 ↑
/ \
5  1
...

Why: We start from the top to find where paths to 5 and 1 diverge

Step 2: Recursively search left subtree for nodes 5 or 1

5 ↑
/ \
6  2
...

Why: Check if left subtree contains either node

Step 3: Found node 5 in left subtree, return node 5 up

Returning node 5 from left subtree

Why: One target found, propagate upwards

Step 4: Recursively search right subtree for nodes 5 or 1

1 ↑
/ \
0  8

Why: Check if right subtree contains either node

Step 5: Found node 1 in right subtree, return node 1 up

Returning node 1 from right subtree

Why: Second target found, propagate upwards

Step 6: At node 3, left returned 5 and right returned 1, so 3 is LCA

3 is lowest common ancestor of 5 and 1

Why: Both targets found in different subtrees, so current node is LCA

Result:

3
Final answer: Lowest Common Ancestor is node 3

Annotated Code

DSA Go

package main

import "fmt"

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

func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
	if root == nil || root == p || root == q {
		return root // base case: found p or q or reached leaf
	}

	left := lowestCommonAncestor(root.Left, p, q) // search left subtree
	right := lowestCommonAncestor(root.Right, p, q) // search right subtree

	if left != nil && right != nil {
		return root // p and q found in different subtrees, root is LCA
	}

	if left != nil {
		return left // p and q both in left subtree
	}

	return right // p and q both in right subtree or none found
}

func main() {
	// Construct the tree
	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.Left.Right.Left = &TreeNode{Val: 7}
	root.Left.Right.Right = &TreeNode{Val: 4}
	root.Right.Left = &TreeNode{Val: 0}
	root.Right.Right = &TreeNode{Val: 8}

	p := root.Left       // node 5
	q := root.Right      // node 1

	lca := lowestCommonAncestor(root, p, q)
	fmt.Printf("Lowest Common Ancestor of %d and %d is %d\n", p.Val, q.Val, lca.Val)
}

if root == nil || root == p || root == q {

stop recursion if current node is null or matches p or q

left := lowestCommonAncestor(root.Left, p, q)

search left subtree for p or q

right := lowestCommonAncestor(root.Right, p, q)

search right subtree for p or q

if left != nil && right != nil {

if p and q found in different subtrees, current node is LCA

if left != nil {

if both found in left subtree, return left

return right

otherwise return right subtree result

Output

Lowest Common Ancestor of 5 and 1 is 3

Complexity Analysis

Time: O(n) because we visit each node once in the worst case

Space: O(h) where h is tree height due to recursion stack

vs Alternative: Compared to storing paths for both nodes and comparing, this approach uses less space and is simpler

Edge Cases

One or both nodes not present in tree

Function returns nil or the node that exists; no false LCA

DSA Go

if root == nil || root == p || root == q {

Both nodes are the same

Returns that node as LCA immediately

DSA Go

if root == nil || root == p || root == q {

Tree with only root node

If p and q are root, returns root; else nil

DSA Go

if root == nil || root == p || root == q {

When to Use This Pattern

When asked to find the closest shared ancestor of two nodes in a tree, use the Lowest Common Ancestor pattern by recursive subtree search to identify where paths split.

Common Mistakes

Mistake: Returning the first found node without checking both subtrees
Fix: Check both left and right subtree results before deciding LCA

Mistake: Not handling the base case when root matches p or q
Fix: Add base case to return root if it matches p or q

Summary

Finds the lowest node in a binary tree that is an ancestor of two given nodes.

Use when you need to find the closest shared parent of two nodes in a tree.

The key insight is that if one node is found in each subtree, the current node is their lowest common ancestor.

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