Given values of two values n1 and n2 in a Binary Search Tree, find the **L**owest **C**ommon **A**ncestor (LCA). You may assume that both the values exist in the tree.

LCA of 10 and 14 is 12 LCA of 14 and 8 is 8 LCA of 10 and 22 is 20

**Following is definition of LCA from Wikipedia:**

Let T be a rooted tree. The lowest common ancestor between two nodes n1 and n2 is defined as the lowest node in T that has both n1 and n2 as descendants (where we allow a node to be a descendant of itself).

The LCA of n1 and n2 in T is the shared ancestor of n1 and n2 that is located farthest from the root. Computation of lowest common ancestors may be useful, for instance, as part of a procedure for determining the distance between pairs of nodes in a tree: the distance from n1 to n2 can be computed as the distance from the root to n1, plus the distance from the root to n2, minus twice the distance from the root to their lowest common ancestor. (Source Wiki)

If we are given a BST where every node has **parent pointer**, then LCA can be easily determined by traversing up using parent pointer and printing the first intersecting node.

We can solve this problem using BST properties. We can **recursively traverse** the BST from root. The main idea of the solution is, while traversing from top to bottom, the first node n we encounter with value between n1 and n2, i.e., n1 < n < n2 or same as one of the n1 or n2, is LCA of n1 and n2 (assuming that n1 < n2). So just recursively traverse the BST in, if node’s value is greater than both n1 and n2 then our LCA lies in left side of the node, if it’s is smaller than both n1 and n2, then LCA lies on right side. Otherwise root is LCA (assuming that both n1 and n2 are present in BST)

## C

`// A recursive C program to find LCA of two nodes n1 and n2. ` `#include <stdio.h> ` `#include <stdlib.h> ` ` ` `struct` `node ` `{ ` ` ` `int` `data; ` ` ` `struct` `node* left, *right; ` `}; ` ` ` `/* Function to find LCA of n1 and n2. The function assumes that both ` ` ` `n1 and n2 are present in BST */` `struct` `node *lca(` `struct` `node* root, ` `int` `n1, ` `int` `n2) ` `{ ` ` ` `if` `(root == NULL) ` `return` `NULL; ` ` ` ` ` `// If both n1 and n2 are smaller than root, then LCA lies in left ` ` ` `if` `(root->data > n1 && root->data > n2) ` ` ` `return` `lca(root->left, n1, n2); ` ` ` ` ` `// If both n1 and n2 are greater than root, then LCA lies in right ` ` ` `if` `(root->data < n1 && root->data < n2) ` ` ` `return` `lca(root->right, n1, n2); ` ` ` ` ` `return` `root; ` `} ` ` ` `/* Helper function that allocates a new node with the given data.*/` `struct` `node* newNode(` `int` `data) ` `{ ` ` ` `struct` `node* node = (` `struct` `node*)` `malloc` `(` `sizeof` `(` `struct` `node)); ` ` ` `node->data = data; ` ` ` `node->left = node->right = NULL; ` ` ` `return` `(node); ` `} ` ` ` `/* Driver program to test lca() */` `int` `main() ` `{ ` ` ` `// Let us construct the BST shown in the above figure ` ` ` `struct` `node *root = newNode(20); ` ` ` `root->left = newNode(8); ` ` ` `root->right = newNode(22); ` ` ` `root->left->left = newNode(4); ` ` ` `root->left->right = newNode(12); ` ` ` `root->left->right->left = newNode(10); ` ` ` `root->left->right->right = newNode(14); ` ` ` ` ` `int` `n1 = 10, n2 = 14; ` ` ` `struct` `node *t = lca(root, n1, n2); ` ` ` `printf` `(` `"LCA of %d and %d is %d \n"` `, n1, n2, t->data); ` ` ` ` ` `n1 = 14, n2 = 8; ` ` ` `t = lca(root, n1, n2); ` ` ` `printf` `(` `"LCA of %d and %d is %d \n"` `, n1, n2, t->data); ` ` ` ` ` `n1 = 10, n2 = 22; ` ` ` `t = lca(root, n1, n2); ` ` ` `printf` `(` `"LCA of %d and %d is %d \n"` `, n1, n2, t->data); ` ` ` ` ` `getchar` `(); ` ` ` `return` `0; ` `} ` |

## Java

`// Recursive Java program to print lca of two nodes ` ` ` `// A binary tree node ` `class` `Node ` `{ ` ` ` `int` `data; ` ` ` `Node left, right; ` ` ` ` ` `Node(` `int` `item) ` ` ` `{ ` ` ` `data = item; ` ` ` `left = right = ` `null` `; ` ` ` `} ` `} ` ` ` `class` `BinaryTree ` `{ ` ` ` `Node root; ` ` ` ` ` `/* Function to find LCA of n1 and n2. The function assumes that both ` ` ` `n1 and n2 are present in BST */` ` ` `Node lca(Node node, ` `int` `n1, ` `int` `n2) ` ` ` `{ ` ` ` `if` `(node == ` `null` `) ` ` ` `return` `null` `; ` ` ` ` ` `// If both n1 and n2 are smaller than root, then LCA lies in left ` ` ` `if` `(node.data > n1 && node.data > n2) ` ` ` `return` `lca(node.left, n1, n2); ` ` ` ` ` `// If both n1 and n2 are greater than root, then LCA lies in right ` ` ` `if` `(node.data < n1 && node.data < n2) ` ` ` `return` `lca(node.right, n1, n2); ` ` ` ` ` `return` `node; ` ` ` `} ` ` ` ` ` `/* Driver program to test lca() */` ` ` `public` `static` `void` `main(String args[]) ` ` ` `{ ` ` ` `// Let us construct the BST shown in the above figure ` ` ` `BinaryTree tree = ` `new` `BinaryTree(); ` ` ` `tree.root = ` `new` `Node(` `20` `); ` ` ` `tree.root.left = ` `new` `Node(` `8` `); ` ` ` `tree.root.right = ` `new` `Node(` `22` `); ` ` ` `tree.root.left.left = ` `new` `Node(` `4` `); ` ` ` `tree.root.left.right = ` `new` `Node(` `12` `); ` ` ` `tree.root.left.right.left = ` `new` `Node(` `10` `); ` ` ` `tree.root.left.right.right = ` `new` `Node(` `14` `); ` ` ` ` ` `int` `n1 = ` `10` `, n2 = ` `14` `; ` ` ` `Node t = tree.lca(tree.root, n1, n2); ` ` ` `System.out.println(` `"LCA of "` `+ n1 + ` `" and "` `+ n2 + ` `" is "` `+ t.data); ` ` ` ` ` `n1 = ` `14` `; ` ` ` `n2 = ` `8` `; ` ` ` `t = tree.lca(tree.root, n1, n2); ` ` ` `System.out.println(` `"LCA of "` `+ n1 + ` `" and "` `+ n2 + ` `" is "` `+ t.data); ` ` ` ` ` `n1 = ` `10` `; ` ` ` `n2 = ` `22` `; ` ` ` `t = tree.lca(tree.root, n1, n2); ` ` ` `System.out.println(` `"LCA of "` `+ n1 + ` `" and "` `+ n2 + ` `" is "` `+ t.data); ` ` ` ` ` `} ` `} ` ` ` `// This code has been contributed by Mayank Jaiswal ` |

## Python

`# A recursive python program to find LCA of two nodes ` `# n1 and n2 ` ` ` `# A Binary tree node ` `class` `Node: ` ` ` ` ` `# Constructor to create a new node ` ` ` `def` `__init__(` `self` `, data): ` ` ` `self` `.data ` `=` `data ` ` ` `self` `.left ` `=` `None` ` ` `self` `.right ` `=` `None` ` ` `# Function to find LCA of n1 and n2. The function assumes ` `# that both n1 and n2 are present in BST ` `def` `lca(root, n1, n2): ` ` ` ` ` `# Base Case ` ` ` `if` `root ` `is` `None` `: ` ` ` `return` `None` ` ` ` ` `# If both n1 and n2 are smaller than root, then LCA ` ` ` `# lies in left ` ` ` `if` `(root.data > n1 ` `and` `root.data > n2): ` ` ` `return` `lca(root.left, n1, n2) ` ` ` ` ` `# If both n1 and n2 are greater than root, then LCA ` ` ` `# lies in right ` ` ` `if` `(root.data < n1 ` `and` `root.data < n2): ` ` ` `return` `lca(root.right, n1, n2) ` ` ` ` ` `return` `root ` ` ` `# Driver program to test above function ` ` ` `# Let us construct the BST shown in the figure ` `root ` `=` `Node(` `20` `) ` `root.left ` `=` `Node(` `8` `) ` `root.right ` `=` `Node(` `22` `) ` `root.left.left ` `=` `Node(` `4` `) ` `root.left.right ` `=` `Node(` `12` `) ` `root.left.right.left ` `=` `Node(` `10` `) ` `root.left.right.right ` `=` `Node(` `14` `) ` ` ` `n1 ` `=` `10` `; n2 ` `=` `14` `t ` `=` `lca(root, n1, n2) ` `print` `"LCA of %d and %d is %d"` `%` `(n1, n2, t.data) ` ` ` `n1 ` `=` `14` `; n2 ` `=` `8` `t ` `=` `lca(root, n1, n2) ` `print` `"LCA of %d and %d is %d"` `%` `(n1, n2 , t.data) ` ` ` `n1 ` `=` `10` `; n2 ` `=` `22` `t ` `=` `lca(root, n1, n2) ` `print` `"LCA of %d and %d is %d"` `%` `(n1, n2, t.data) ` ` ` `# This code is contributed by Nikhil Kumar Singh(nickzuck_007) ` |

Output:

LCA of 10 and 14 is 12 LCA of 14 and 8 is 8 LCA of 10 and 22 is 20

Time complexity of above solution is O(h) where h is height of tree. Also, the above solution requires O(h) extra space in function call stack for recursive function calls. We can avoid extra space using **iterative solution**.

`/* Function to find LCA of n1 and n2. The function assumes that both ` ` ` `n1 and n2 are present in BST */` `struct` `node *lca(` `struct` `node* root, ` `int` `n1, ` `int` `n2) ` `{ ` ` ` `while` `(root != NULL) ` ` ` `{ ` ` ` `// If both n1 and n2 are smaller than root, then LCA lies in left ` ` ` `if` `(root->data > n1 && root->data > n2) ` ` ` `root = root->left; ` ` ` ` ` `// If both n1 and n2 are greater than root, then LCA lies in right ` ` ` `else` `if` `(root->data < n1 && root->data < n2) ` ` ` `root = root->right; ` ` ` ` ` `else` `break` `; ` ` ` `} ` ` ` `return` `root; ` `} ` |

See this for complete program.

You may like to see below articles as well :

Lowest Common Ancestor in a Binary Tree

Find LCA in Binary Tree using RMQ

**Exercise**

The above functions assume that n1 and n2 both are in BST. If n1 and n2 are not present, then they may return incorrect result. Extend the above solutions to return NULL if n1 or n2 or both not present in BST.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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