What if you could find the closest common ancestor in a family tree instantly, no matter how big it is?
Why Lowest Common Ancestor in Binary Tree in DSA Java?
Imagine you have a family tree on paper and want to find the closest common grandparent of two cousins. You try to trace each cousin's path up to the root manually, comparing each step.
This manual tracing is slow and confusing. You might miss a step or compare wrong ancestors. It becomes very hard when the tree is big or unbalanced, and you waste time checking the same nodes multiple times.
The Lowest Common Ancestor (LCA) method quickly finds the closest shared ancestor of two nodes in a binary tree by smartly exploring the tree only once. It avoids repeated checks and gives the answer efficiently.
Node ancestor1 = node1; while (!isAncestor(ancestor1, node1) || !isAncestor(ancestor1, node2)) { ancestor1 = ancestor1.parent; } return ancestor1;
Node findLCA(Node root, Node node1, Node node2) {
if (root == null || 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 != null ? left : right;
}This concept enables fast and reliable discovery of shared ancestors in trees, which is crucial for many problems like genealogy, file systems, and network routing.
In a company's organizational chart, finding the lowest common manager of two employees helps decide who should approve joint projects or resolve conflicts.
Manual ancestor search is slow and error-prone.
LCA uses a single tree traversal to find the closest shared ancestor.
This method works efficiently even for large and complex 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: Clarify the options
The deepest node that is an ancestor of both nodes correctly describes the LCA. The root node is not always the LCA, and the smallest value or child relationships do not define LCA.Final Answer:
The deepest node that is an ancestor of both nodes -> Option BQuick Check:
LCA = deepest common ancestor [OK]
- Confusing LCA with root node
- Thinking LCA is the smaller value node
- Assuming LCA is a child of given nodes
Solution
Step 1: Analyze method parameters and return type
The method should take the root node and two node values (or nodes) to find their LCA. It returns the TreeNode representing the LCA.Step 2: Match with options
public TreeNode findLCA(TreeNode root, int n1, int n2) matches this: it takes root and two integers, returns TreeNode. public int findLCA(int root, TreeNode n1, TreeNode n2) has wrong parameter types, C misses root, D returns void which is incorrect.Final Answer:
public TreeNode findLCA(TreeNode root, int n1, int n2) -> Option DQuick Check:
Correct signature includes root and returns TreeNode [OK]
- Omitting root parameter
- Wrong return type (void or int)
- Incorrect parameter order or types
3 / \ 5 1 / \ / \ 6 2 0 8 / \ 7 4 TreeNode lca = findLCA(root, 5, 1); System.out.println(lca.val);
Solution
Step 1: Identify 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 their lowest common ancestor
The lowest common ancestor of 5 and 1 is the root node 3, as it is the deepest node that has both as descendants.Final Answer:
3 -> Option AQuick Check:
LCA(5,1) = 3 [OK]
- Choosing one of the nodes instead of their ancestor
- Confusing child nodes as LCA
- Ignoring tree structure
public TreeNode findLCA(TreeNode root, int n1, int n2) {
if (root == null) return null;
if (root.val == n1 || root.val == n2) return root;
TreeNode left = findLCA(root.left, n1, n2);
TreeNode right = findLCA(root.right, n1, n2);
if (left != null && right != null) return root;
if (left != null) return left;
return right;
}Solution
Step 1: Understand the method logic
The method returns root if it matches either node, then recursively searches left and right subtrees. If both sides return non-null, root is LCA.Step 2: Identify missing case
The method assumes both nodes exist. If one node is missing, it may return the other node incorrectly as LCA without confirming presence of both.Final Answer:
It does not handle the case when one node is not present in the tree -> Option CQuick Check:
Missing node presence check causes incorrect LCA [OK]
- Ignoring missing node cases
- Assuming both nodes always exist
- Not validating subtree results properly
Solution
Step 1: Understand the problem with duplicates
When values are duplicated, comparing node values is unreliable to identify nodes uniquely.Step 2: Use node references for comparison
Instead of comparing values, compare the actual node references (memory addresses) to find exact nodes during recursion.Step 3: Adjust algorithm accordingly
Modify the LCA method to check if current node == target node references, not values.Final Answer:
Use node references directly instead of comparing values during recursion -> Option AQuick Check:
Use references, not values, for duplicates [OK]
- Comparing values instead of references
- Trying to remove duplicates unnecessarily
- Using value-based hash maps causing errors
Ready to prove it?
Pass the test and this pattern moves unknown
