Given the root of a binary search tree and two nodes n1 and n2, find the LCA(Lowest Common Ancestor) of the nodes in a given binary search tree.

Table of Contents

## Example

## Naive Approach for Lowest Common Ancestor in Binary Search Tree

Find the LCA(n1, n2) using the optimal approach to find LCA in the Binary tree, as BST is also a binary tree.

If we assume that n1 and n2, for which we have to find LCA, exists in the tree, then the above problem can be solved in a single traversal.

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Traverse the tree, for every node we have one of the four cases,

- The current node is either n1 or n2, in this case, we return the node.
- One of the subtrees of the current node contains n1 and the other contains n2, this node is the LCA, return the node.
- One subtree of the current node contains both n1 and n2, we return what the subtree returns.
- None of the subtrees contains n1 and n2, return null.

Time Complexity = **O(n), with single traversal **

where n is the number of nodes in the tree.

## Optimal Approach for Lowest Common Ancestor in Binary Search Tree

Using the properties of BST, the LCA can be found in much lesser time complexity.

- Traverse the BST starting from the root (Initialize curr as root).
- If the current node’s value lies between n1 and n2(both inclusive), then this is the LCA.
- Else if node’s value is less than both n1 and n2, LCA is present in the right half (curr becomes curr.right).
- Else LCA is present in the left half (curr becomes curr.left).

## Explanation

Consider the BST in the above image, lets find the LCA(32, 40)

Start traversing the tree starting from the root.

- Current node’s value = 20

20 < 32 and 20 < 40, LCA is present in right sub tree - Current node’s value = 24

24 < 32 and 20 < 40, LCA is present in right sub tre - Current node’s value = 35

35 >= 32 and 35 <= 40, so this is the LCA

That is,**LCA(32, 40) = 35**

## JAVA Code for Lowest Common Ancestor in Binary Search Tree

public class LCABST { // Class to represent a node in BST static class Node { int data; Node left, right; public Node(int data) { this.data = data; left = right = null; } } // Function to return LCA of two nodes, and return -1 if LCA does not exist private static int LCA(Node root, int n1, int n2) { // No LCA for an empty tree if (root == null) { return -1; } // Traverse the tree starting from root Node curr = root; int lca = -1; while (curr != null) { if (curr.data < n1 && curr.data < n2) { // LCA is present in the right sub tree curr = curr.right; } else if (curr.data > n1 && curr.data > n2) { // LCA is present in left sub tree curr = curr.left; } else { lca = curr.data; break; } } // Return LCA return lca; } public static void main(String[] args) { // Constructing the BST in above example Node root = new Node(20); root.left = new Node(11); root.right = new Node(24); root.right.left = new Node(21); root.right.right = new Node(35); root.right.right.left = new Node(32); root.right.right.right = new Node(40); // Queries System.out.println(LCA(root, 24, 40)); System.out.println(LCA(root, 11, 21)); System.out.println(LCA(root, 32, 40)); } }

## C++ Code for Lowest Common Ancestor in Binary Search Tree

#include <iostream> using namespace std; // Class representing node of binary search tree class Node { public: int data; Node *left; Node *right; }; // Function to allocate new memory to a tree node Node* newNode(int data) { Node *node = new Node(); node->data = data; node->left = NULL; node->right = NULL; return (node); } // Function to return LCA of two nodes, and return -1 if LCA does not exist int LCA(Node *root, int n1, int n2) { // No LCA for an empty tree if (root == NULL) { return -1; } // Traverse the tree starting from root Node *curr = root; int lca = -1; while (curr != NULL) { if (curr->data < n1 && curr->data < n2) { // LCA is present in the right sub tree curr = curr->right; } else if (curr->data > n1 && curr->data > n2) { // LCA is present in left sub tree curr = curr->left; } else { lca = curr->data; break; } } // Return LCA return lca; } int main() { // Construct the tree shown in above example Node *root = newNode(20); root->left = newNode(11); root->right = newNode(24); root->right->left = newNode(21); root->right->right = newNode(35); root->right->right->left = newNode(32); root->right->right->right = newNode(40); // Queries cout<<LCA(root, 24, 40)<<endl; cout<<LCA(root, 11, 21)<<endl; cout<<LCA(root, 32, 40)<<endl; return 0; }

24 20 35

## Complexity Analysis

Time Complexity = **O(h), **where h is the height of BST