What if you could find the closest common boss of two employees instantly, no matter how big the company?
Why Lowest Common Ancestor in Binary Tree in DSA Go?
Imagine you have a family tree drawn on paper, and you want to find the closest common grandparent of two cousins. You try to trace each cousin's path back to the top manually, comparing each step. This gets confusing and slow as the tree grows bigger.
Manually tracing paths for each node and comparing them is slow and error-prone. You might miss a step or get lost in the branches. It's like trying to find a meeting point in a maze without a map.
The Lowest Common Ancestor (LCA) algorithm quickly finds the closest shared ancestor of two nodes in a binary tree by exploring the tree just once. It acts like a smart guide that knows exactly where to look, saving time and avoiding mistakes.
func findPath(root *Node, target int, path *[]*Node) bool {
if root == nil {
return false
}
*path = append(*path, root)
if root.value == target {
return true
}
if findPath(root.left, target, path) || findPath(root.right, target, path) {
return true
}
*path = (*path)[:len(*path)-1]
return false
}
// Then compare paths for two nodesfunc lowestCommonAncestor(root, p, q *Node) *Node {
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 lets you instantly find the closest shared ancestor of any two nodes in a tree, enabling fast queries in family trees, file systems, and network structures.
In a company's organizational chart, finding the lowest common manager of two employees helps decide who should approve a joint project.
Manual path tracing is slow and confusing for large trees.
LCA algorithm finds the closest shared ancestor efficiently with one tree traversal.
Useful in many hierarchical data problems like family trees and org charts.
Practice
Solution
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).Step 2: Identify the deepest common ancestor
Among all common ancestors, the LCA is the one farthest from the root (deepest).Final Answer:
The deepest node that is an ancestor of both nodes -> Option CQuick Check:
LCA = deepest common ancestor [OK]
- Confusing LCA with root node always
- Thinking LCA is the node with smaller value
- Assuming LCA is a child of both nodes
Solution
Step 1: Understand pointer usage in Go for tree nodes
Tree nodes are usually passed as pointers to avoid copying and to allow modification.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.Final Answer:
func lowestCommonAncestor(root *TreeNode, p *TreeNode, q *TreeNode) *TreeNode -> Option DQuick Check:
Use pointers for nodes in Go function signature [OK]
- Passing nodes by value instead of pointer
- Mixing pointer and value types in parameters
- Returning value instead of pointer
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)
}Solution
Step 1: Trace recursive calls for nodes 5 and 1
Starting at root (3), left subtree contains node 5, right subtree contains node 1.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.Final Answer:
3 -> Option AQuick Check:
LCA(5,1) = 3 [OK]
- Returning one of the nodes instead of root
- Confusing node values with LCA
- Not handling nil checks properly
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
}Solution
Step 1: Analyze return conditions after recursion
If both left and right are nil, the function incorrectly returns root instead of nil.Step 2: Understand correct base case for no ancestor found
When neither subtree contains p or q, function should return nil, not root.Final Answer:
Returns root when both left and right are nil instead of nil -> Option AQuick Check:
Return nil if no ancestor found [OK]
- Returning root instead of nil when no match
- Missing nil check at start
- Comparing values instead of pointers incorrectly
p and q?Solution
Step 1: Understand problem with duplicate values
When values are duplicated, comparing values alone can confuse nodes.Step 2: Use node pointers to uniquely identify nodes
Comparing pointers ensures the exact nodes p and q are found, not just matching values.Step 3: Modify function to compare pointers (e.g., root == p)
This guarantees correct LCA even with duplicate values.Final Answer:
Compare node pointers instead of node values to identify nodes -> Option BQuick Check:
Use pointers, not values, for unique node identity [OK]
- Comparing only values causing wrong LCA
- Assuming duplicates don't affect LCA
- Using parent pointers unnecessarily
Ready to prove it?
Pass the test and this pattern moves unknown
