Binary Search Tree : Lowest Common Ancestor HackerRank Solution

Binary Search Tree : Lowest Common Ancestor HackerRank Solution
Source : https://www.hackerrank.com/challenges/binary-search-tree-lowest-common-ancestor

You are given pointer to the root of the binary search tree and two values and . You need to return the lowest common ancestor (LCA) of and in the binary search tree.

In the diagram above, the lowest common ancestor of the nodes and is the node . Node is the lowest node which has nodes and as descendants.

Function Description

Complete the function lca in the editor below. It should return a pointer to the lowest common ancestor node of the two values given.

lca has the following parameters:
- root: a pointer to the root node of a binary search tree
- v1: a node.data value
- v2: a node.data value

Input Format

The first line contains an integer, , the number of nodes in the tree.
The second line contains space-separated integers representing values.
The third line contains two space-separated integers, and .

To use the test data, you will have to create the binary search tree yourself. Here on the platform, the tree will be created for you.

Constraints

The tree will contain nodes with data equal to and .

Output Format

Return the a pointer to the node that is the lowest common ancestor of and .

Sample Input

64 2 3 1 7 61 7

and .

Sample Output

[reference to node 4]

Explanation

LCA of and is , the root in this case.
Return a pointer to the node.

Source : https://www.hackerrank.com/challenges/binary-search-tree-lowest-common-ancestor

Solution

	// Karthikalapati.blogspot.com
	/* Node is defined as :
	class Node {
	int data;
	Node left;
	Node right;
	}
	*/
	// Assumes tree has unique values.
	// This problem is simpler since we're given a binary SEARCH tree.
	// Time Complexity: O(log n) on a balanced tree
	// Space Complexity: O(1)
	static Node lca(Node n, int v1, int v2) {
	while (n != null) {
	if (n.data > v1 && n.data > v2) {
	n = n.left;
	} else if (n.data < v1 && n.data < v2) {
	n = n.right;
	} else {
	break;
	}
	}
	return n;
	}