Lowest Common Ancestor in Binary Tree in DSA Go - Build from Scratch

Lowest Common Ancestor in Binary Tree

📖 Scenario: You are working on a family tree application. You want to find the closest common ancestor of two family members in the tree.

🎯 Goal: Build a program that finds the Lowest Common Ancestor (LCA) of two nodes in a binary tree.

📋 What You'll Learn

Create a binary tree with specific nodes

Add variables for the two target nodes

Implement a function to find the LCA

Print the value of the LCA node

💡 Why This Matters

🌍 Real World

Finding the lowest common ancestor is useful in family trees, organizational charts, and network routing to find shared connections.

💼 Career

Understanding tree traversal and recursive algorithms is important for software engineering roles involving data structures, databases, and system design.

1

Create the Binary Tree

Create a binary tree with the following structure using TreeNode structs:

Root node with value 3
Left child with value 5
Right child with value 1
Left child of node 5 with value 6
Right child of node 5 with value 2
Left child of node 1 with value 0
Right child of node 1 with value 8
Left child of node 2 with value 7
Right child of node 2 with value 4
Assign the root node to a variable called root.

DSA Go

package main

import "fmt"

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

func main() {
    // Create the binary tree nodes and link them
    // Your code here
}

Hint

Start by creating the root node with value 3. Then create child nodes and assign them to the Left and Right fields accordingly.

2

Set Target Nodes for LCA

Create two variables called p and q that point to the nodes with values 5 and 1 respectively in the tree.

DSA Go

package main

import "fmt"

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

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}
    root.Left.Right.Left = &TreeNode{Val: 7}
    root.Left.Right.Right = &TreeNode{Val: 4}

    // Create variables p and q for nodes with values 5 and 1
    // Your code here
}

Hint

Assign p to the left child of root and q to the right child of root.

3

Implement the Lowest Common Ancestor Function

Write a function called lowestCommonAncestor that takes root, p, and q as parameters (all of type *TreeNode) and returns the lowest common ancestor node. Use recursion to find the LCA.

DSA Go

package main

import "fmt"

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

func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
    // Implement the recursive LCA logic here
    // Your code here
}

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}
    root.Left.Right.Left = &TreeNode{Val: 7}
    root.Left.Right.Right = &TreeNode{Val: 4}

    p := root.Left
    q := root.Right
}

Hint

Use recursion to search left and right subtrees. If both sides return non-nil, current node is LCA. If one side returns non-nil, return that side.

4

Print the Lowest Common Ancestor Value

Call the lowestCommonAncestor function with root, p, and q. Print the Val field of the returned node using fmt.Println.

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
    }
    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}
    root.Left.Right.Left = &TreeNode{Val: 7}
    root.Left.Right.Right = &TreeNode{Val: 4}

    p := root.Left
    q := root.Right

    // Print the LCA value
    // Your code here
}

Hint

Call lowestCommonAncestor(root, p, q) and store the result in a variable. Then print the Val of that variable.

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