What if you could find the closest common ancestor in a family tree with just one smart search?
Why Lowest Common Ancestor in Binary Tree in DSA C#?
Imagine you have a family tree drawn on paper, and you want to find the closest common grandparent of two cousins. Doing this by tracing each person's ancestors manually can be confusing and slow, especially if the tree is big.
Manually checking each ancestor for both cousins means going back and forth repeatedly. It's easy to miss a connection or get lost in the branches. This wastes time and can lead to mistakes.
The Lowest Common Ancestor (LCA) method helps you find the closest shared ancestor quickly by exploring the tree just once. It uses a smart way to check nodes, so you don't have to trace every path separately.
bool IsAncestor(Node root, Node target) {
if (root == null) return false;
if (root == target) return true;
return IsAncestor(root.left, target) || IsAncestor(root.right, target);
}
Node FindLCA(Node root, Node node1, Node node2) {
if (!IsAncestor(root, node1) || !IsAncestor(root, node2)) return null;
// Manually check each node
// ...
return null; // slow and complex
}Node FindLCA(Node root, Node node1, Node node2) {
if (root == null) return null;
if (root == node1 || root == node2) return root;
Node left = FindLCA(root.left, node1, node2);
Node right = FindLCA(root.right, node1, node2);
if (left != null && right != null) return root;
return left ?? right;
}This lets you quickly find the closest shared ancestor of any two nodes in a tree, making complex family or organizational relationships easy to understand.
In a company's organizational chart, finding the lowest common manager of two employees helps decide who should approve a joint project.
Manual ancestor checks are slow and error-prone.
LCA finds the closest shared ancestor efficiently with one tree traversal.
This method simplifies understanding relationships in trees.
Practice
Solution
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 (a node can be a descendant of itself).Step 2: Identify the correct description
Among the options, only the deepest common ancestor fits this definition.Final Answer:
The deepest node that is an ancestor of both nodes -> Option DQuick Check:
LCA = deepest common ancestor [OK]
- Confusing LCA with root node
- Thinking LCA is the smaller value node
- Assuming LCA is a child of both nodes
TreeNode?Solution
Step 1: Analyze method parameters and return type
The method should accept the root node and two nodes to find LCA for, and return a TreeNode representing the LCA.Step 2: Match signature with correct types
public TreeNode FindLCA(TreeNode root, TreeNode p, TreeNode q) correctly uses TreeNode for all nodes and returns TreeNode. Other options have incorrect parameter types or return types.Final Answer:
public TreeNode FindLCA(TreeNode root, TreeNode p, TreeNode q) -> Option AQuick Check:
Correct signature uses TreeNode parameters and return type [OK]
- Using int instead of TreeNode for nodes
- Wrong return type (void or int)
- Incorrect access modifier or parameter order
FindLCA(root, node5, node1).val?
3 / \ 5 1 / \ / \ 6 2 0 8 / \ 7 4 // Code snippet: TreeNode lca = FindLCA(root, node5, node1); Console.WriteLine(lca.val);
Solution
Step 1: Locate nodes 5 and 1 in the tree
Node 5 is left child of root 3, node 1 is right child of root 3.Step 2: Find lowest common ancestor
The root node 3 is the lowest node that has both 5 and 1 as descendants.Final Answer:
3 -> Option BQuick Check:
LCA(5,1) = 3 [OK]
- Choosing one of the nodes instead of their ancestor
- Confusing node values with LCA
- Ignoring tree structure
public TreeNode FindLCA(TreeNode root, TreeNode p, TreeNode q) {
if (root == null) return null;
if (root == p || root == q) return root;
TreeNode left = FindLCA(root.left, p, q);
TreeNode right = FindLCA(root.right, p, q);
if (left != null && right != null) return root;
return left != null ? left : right;
}Solution
Step 1: Review base cases and recursive calls
The method returns null if root is null, returns root if root matches p or q, then recursively searches left and right subtrees.Step 2: Check logic for combining results
If both left and right are non-null, root is LCA; else returns whichever subtree found a node. This is correct logic.Final Answer:
No bug, method is correct -> Option CQuick Check:
Standard LCA recursive method is correct [OK]
- Comparing node values instead of references if nodes are unique
- Missing base case for null root
- Incorrect return logic
Solution
Step 1: Understand the problem with non-unique values
Since values are not unique, we cannot rely on node references or values alone to find LCA directly.Step 2: Use path comparison approach
Find paths from root to each node value (may find multiple nodes), then compare paths to find the last common node, which is the LCA.Final Answer:
Use a method that finds paths from root to each node value and then compare paths to find LCA -> Option AQuick Check:
Path comparison works for non-unique values [OK]
- Assuming node references are unique
- Returning root always
- Using hash set without paths
Ready to prove it?
Pass the test and this pattern moves unknown
