# Lowest Common Ancestor in a Binary Search Tree.

Given two values n1 and n2 in a Binary Search Tree, find the Lowest Common Ancestor (LCA). You may assume that both values exist in the tree.

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 [i.e., closest to n1 and n2].

Examples:

Input Tree:

Input: LCA of 10 and 14
Output:  12
Explanation: 12 is the closest node to both 10 and 14
which is a ancestor of both the nodes.

Input: LCA of 8 and 14
Output:  8
Explanation: 8 is the closest node to both 8 and 14
which is a ancestor of both the nodes.

## Lowest Common Ancestor in a Binary Search Tree using Recursion:

To solve the problem follow the below idea:

For Binary search tree, while traversing the tree from top to bottom the first node which lies in between the two numbers n1 and n2 is the LCA of the nodes, i.e. the first node n with the lowest depth which lies in between n1 and n2 (n1<=n<=n2) n1 < n2.

So just recursively traverse the BST , if node’s value is greater than both n1 and n2 then our LCA lies in the left side of the node, if it is smaller than both n1 and n2, then LCA lies on the right side. Otherwise, the root is LCA (assuming that both n1 and n2 are present in BST)

Follow the given steps to solve the problem:

• Create a recursive function that takes a node and the two values n1 and n2.
• If the value of the current node is less than both n1 and n2, then LCA lies in the right subtree. Call the recursive function for the right subtree.
• If the value of the current node is greater than both n1 and n2, then LCA lies in the left subtree. Call the recursive function for the left subtree.
• If both the above cases are false then return the current node as LCA.

Below is the implementation of the above approach.

## C++

 `// A recursive CPP program to find` `// LCA of two nodes n1 and n2.` `#include ` `using` `namespace` `std;`   `class` `node {` `public``:` `    ``int` `data;` `    ``node *left, *right;` `};`   `/* Function to find LCA of n1 and n2.` `The function assumes that both` `n1 and n2 are present in BST */` `node* lca(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.*/` `node* newNode(``int` `data)` `{` `    ``node* Node = ``new` `node();` `    ``Node->data = data;` `    ``Node->left = Node->right = NULL;` `    ``return` `(Node);` `}`   `/* Driver code*/` `int` `main()` `{` `    ``// Let us construct the BST` `    ``// shown in the above figure` `    ``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);`   `      ``// Function calls` `    ``int` `n1 = 10, n2 = 14;` `    ``node* t = lca(root, n1, n2);` `    ``cout << ``"LCA of "` `<< n1 << ``" and "` `<< n2 << ``" is "` `         ``<< t->data << endl;`   `    ``n1 = 14, n2 = 8;` `    ``t = lca(root, n1, n2);` `    ``cout << ``"LCA of "` `<< n1 << ``" and "` `<< n2 << ``" is "` `         ``<< t->data << endl;`   `    ``n1 = 10, n2 = 22;` `    ``t = lca(root, n1, n2);` `    ``cout << ``"LCA of "` `<< n1 << ``" and "` `<< n2 << ``" is "` `         ``<< t->data << endl;` `    ``return` `0;` `}`   `// This code is contributed by rathbhupendra`

## C

 `// A recursive C program to find LCA of two nodes n1 and n2.` `#include ` `#include `   `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 code */` `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);`   `      ``// Function calls` `    ``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 code */` `    ``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``);`   `          ``// Function calls` `        ``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`

## Python3

 `# 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`     `# Driver code` `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``)`     `# Function calls` `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)`

## C#

 `// Recursive C# program to print lca of two nodes`   `using` `System;`   `// Recursive C#  program to print lca of two nodes`   `// A binary tree node` `public` `class` `Node {` `    ``public` `int` `data;` `    ``public` `Node left, right;`   `    ``public` `Node(``int` `item)` `    ``{` `        ``data = item;` `        ``left = right = ``null``;` `    ``}` `}`   `public` `class` `BinaryTree {` `    ``public` `Node root;`   `    ``/* Function to find LCA of n1 and n2. The function` `       ``assumes that both n1 and n2 are present in BST */` `    ``public` `virtual` `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 code */` `    ``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);`   `          ``// Function calls` `        ``int` `n1 = 10, n2 = 14;` `        ``Node t = tree.lca(tree.root, n1, n2);` `        ``Console.WriteLine(``"LCA of "` `+ n1 + ``" and "` `+ n2` `                          ``+ ``" is "` `+ t.data);`   `        ``n1 = 14;` `        ``n2 = 8;` `        ``t = tree.lca(tree.root, n1, n2);` `        ``Console.WriteLine(``"LCA of "` `+ n1 + ``" and "` `+ n2` `                          ``+ ``" is "` `+ t.data);`   `        ``n1 = 10;` `        ``n2 = 22;` `        ``t = tree.lca(tree.root, n1, n2);` `        ``Console.WriteLine(``"LCA of "` `+ n1 + ``" and "` `+ n2` `                          ``+ ``" is "` `+ t.data);` `    ``}` `}`   `//  This code is contributed by Shrikant13`

## Javascript

 ``

Output

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

```

Time Complexity: O(H). where H is the height of the tree.
Auxiliary Space: O(H), If recursive stack space is ignored, the space complexity of the above solution is constant.

Below is the iterative implementation of the above approach:

## C++

 `// A recursive CPP program to find` `// LCA of two nodes n1 and n2.` `#include ` `using` `namespace` `std;`   `class` `node {` `public``:` `    ``int` `data;` `    ``node *left, *right;` `};`   `/* Function to find LCA of n1 and n2.` `The function assumes that both n1 and n2` `are present in BST */` `node* lca(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;` `}`   `/* Helper function that allocates` `a new node with the given data.*/` `node* newNode(``int` `data)` `{` `    ``node* Node = ``new` `node();` `    ``Node->data = data;` `    ``Node->left = Node->right = NULL;` `    ``return` `(Node);` `}`   `/* Driver code*/` `int` `main()` `{` `    ``// Let us construct the BST` `    ``// shown in the above figure` `    ``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);`   `      ``// Function calls` `    ``int` `n1 = 10, n2 = 14;` `    ``node* t = lca(root, n1, n2);` `    ``cout << ``"LCA of "` `<< n1 << ``" and "` `<< n2 << ``" is "` `         ``<< t->data << endl;`   `    ``n1 = 14, n2 = 8;` `    ``t = lca(root, n1, n2);` `    ``cout << ``"LCA of "` `<< n1 << ``" and "` `<< n2 << ``" is "` `         ``<< t->data << endl;`   `    ``n1 = 10, n2 = 22;` `    ``t = lca(root, n1, n2);` `    ``cout << ``"LCA of "` `<< n1 << ``" and "` `<< n2 << ``" is "` `         ``<< t->data << endl;` `    ``return` `0;` `}`   `// This code is contributed by rathbhupendra`

## C

 `// A recursive C program to find LCA of two nodes n1 and n2.` `#include ` `#include `   `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)` `{` `    ``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;` `}`   `/* 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 code */` `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);`   `    ``// Function calls      ` `    ``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 */` `    ``static` `Node lca(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;` `    ``}`   `    ``/* Driver code */` `    ``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``);`   `          ``// Function calls` `        ``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 is contributed by SHUBHAMSINGH10`

## Python3

 `# 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):` `    ``while` `root:` `        ``# If both n1 and n2 are smaller than root,` `        ``# then LCA lies in left` `        ``if` `root.data > n1 ``and` `root.data > n2:` `            ``root ``=` `root.left`   `        ``# If both n1 and n2 are greater than root,` `        ``# then LCA lies in right` `        ``elif` `root.data < n1 ``and` `root.data < n2:` `            ``root ``=` `root.right`   `        ``else``:` `            ``break`   `    ``return` `root`     `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` `  ``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``)`   `  ``# Function calls` `  ``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 Sumit Bhardwaj (Timus)`

## C#

 `// Recursive C# program to print lca of two nodes`   `using` `System;`   `// Recursive C# program to print lca of two nodes`   `// A binary tree node` `public` `class` `Node {` `    ``public` `int` `data;` `    ``public` `Node left, right;`   `    ``public` `Node(``int` `item)` `    ``{` `        ``data = item;` `        ``left = right = ``null``;` `    ``}` `}`   `public` `class` `BinaryTree {` `    ``public` `Node root;`   `    ``/* Function to find LCA of n1 and n2.` `    ``The function assumes that both` `    ``n1 and n2 are present in BST */` `    ``public` `virtual` `Node lca(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;` `    ``}`   `    ``/* Driver code */` `    ``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);`   `          ``// Function calls` `        ``int` `n1 = 10, n2 = 14;` `        ``Node t = tree.lca(tree.root, n1, n2);` `        ``Console.WriteLine(``"LCA of "` `+ n1 + ``" and "` `+ n2` `                          ``+ ``" is "` `+ t.data);`   `        ``n1 = 14;` `        ``n2 = 8;` `        ``t = tree.lca(tree.root, n1, n2);` `        ``Console.WriteLine(``"LCA of "` `+ n1 + ``" and "` `+ n2` `                          ``+ ``" is "` `+ t.data);`   `        ``n1 = 10;` `        ``n2 = 22;` `        ``t = tree.lca(tree.root, n1, n2);` `        ``Console.WriteLine(``"LCA of "` `+ n1 + ``" and "` `+ n2` `                          ``+ ``" is "` `+ t.data);` `    ``}` `}`   `// This code is contributed by Shrikant13`

## Javascript

 ``

Output

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

```

Time Complexity: O(H). where H is the height of the tree
Auxiliary Space: O(1). The space complexity of the above solution is constant.

### Lowest Common Ancestor in a Binary Search Tree using Morris traversal:

Follow the steps to implement the above approach:

1. Initialize a pointer curr to the root of the tree.
2. While curr is not NULL, do the following:
3. If curr has no left child, check if curr is either of the two nodes we are interested in. If it is, return curr. Otherwise, move curr to its right child.
4. If curr has a left child, find the inorder predecessor pre of curr by moving to the rightmost node in the left subtree of curr.
5. If the right child of pre is NULL, set it to curr and move curr to its left child.
6. If the right child of pre is curr, set it to NULL and restore the original tree structure. Then check if curr is either of the two nodes we are interested in. If it is, return curr. Otherwise, move curr to its right child.
7. If the two nodes we are interested in are not found during the traversal, return NULL

Below is the implementation of the above approach:

## C++

 `// C++ code to implement the morris traversal approach` `#include` `using` `namespace` `std;`   `struct` `TreeNode {` `    ``int` `val;` `    ``TreeNode *left, *right;` `    ``TreeNode(``int` `x) : val(x), left(NULL), right(NULL) {}` `};`   `TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {` `    ``TreeNode* curr = root;` `    ``while` `(curr != NULL) {` `        ``if` `(curr->left == NULL) { ``// if the left subtree is empty, move to the right subtree` `            ``if` `(curr == p || curr == q) ``return` `curr;` `            ``curr = curr->right;` `        ``}` `        ``else` `{` `            ``TreeNode* pre = curr->left;` `            ``while` `(pre->right != NULL && pre->right != curr) pre = pre->right; ``// find the inorder predecessor of the current node` `            ``if` `(pre->right == NULL) { ``// if the predecessor's right child is NULL, make it point to the current node` `                ``pre->right = curr;` `                ``curr = curr->left;` `            ``}` `            ``else` `{ ``// if the predecessor's right child is the current node, restore the tree structure and move to the right subtree` `                ``pre->right = NULL;` `                ``if` `(curr == p || curr == q) ``return` `curr;` `                ``curr = curr->right;` `            ``}` `        ``}` `    ``}` `    ``return` `NULL;` `}` `// Driver Code` `int` `main() {` `    ``/*` `    ``Input Tree:` `              ``5` `           ``/    \` `         ``4       6` `          ``\       \` `           ``3       7` `                    ``\` `                     ``8` `    ``*/` `    ``TreeNode *root = ``new` `TreeNode(5);` `    ``root->left = ``new` `TreeNode(4);` `    ``root->left->right = ``new` `TreeNode(3);` `    ``root->right = ``new` `TreeNode(6);` `    ``root->right->right = ``new` `TreeNode(7);` `    ``root->right->right->right = ``new` `TreeNode(8);`   `    ``TreeNode *p = root->left;` `    ``TreeNode *q = root->left->right;`   `    ``TreeNode *lca1 = lowestCommonAncestor(root, p, q);` `    ``cout <<``" LCA of "``<val<<``" and "``<val<<``" is "``<< lca1->val << endl; ` `    `  `     ``TreeNode  *x = root->right->right;` `     ``TreeNode *y = root->right->right->right;`   `    ``TreeNode *lca2 = lowestCommonAncestor(root, x, y);` `    ``cout <<``" LCA of "``<val<<``" and "``<val<<``" is "``<< lca2->val << endl;`   `    ``return` `0;` `}` `// This code is contributed by Veerendra_Singh_Rajpoot`

## Java

 `/*package whatever //do not write package name here */` `import` `java.io.*;`   `// Define the TreeNode class ` `// representing a node in a binary tree.` `class` `TreeNode {` `    ``int` `val;` `    ``TreeNode left, right;`   `    ``public` `TreeNode(``int` `x) {` `        ``val = x;` `        ``left = ``null``;` `        ``right = ``null``;` `    ``}` `}`   `// Define the Solution class ` `// containing the lowestCommonAncestor method.` `class` `Solution {` `    ``// Function to find the lowest common ancestor (LCA)` `    ``// of two nodes in a binary tree.` `    ``public` `TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {` `        ``TreeNode curr = root;` `        ``while` `(curr != ``null``) {` `            ``// If there is no left child, ` `            ``// traverse to the right child.` `            ``if` `(curr.left == ``null``) {` `                ``// Check if the current node is one of the target nodes.` `                ``if` `(curr == p || curr == q)` `                    ``return` `curr;` `                ``curr = curr.right;` `            ``} ``else` `{` `                ``// If there is a left child, find the in-order predecessor.` `                ``TreeNode pre = curr.left;` `                ``while` `(pre.right != ``null` `&& pre.right != curr)` `                    ``pre = pre.right;` `                ``if` `(pre.right == ``null``) {` `                    ``// Set the right child of the in-order ` `                    ``// predecessor to the current node.` `                    ``pre.right = curr;` `                    ``curr = curr.left;` `                ``} ``else` `{` `                    ``// Remove the link from the in-order ` `                    ``// predecessor to the current node.` `                    ``pre.right = ``null``;` `                    ``// Check if the current node is one of the target nodes.` `                    ``if` `(curr == p || curr == q)` `                        ``return` `curr;` `                    ``curr = curr.right;` `                ``}` `            ``}` `        ``}` `        ``return` `null``;` `    ``}` `}`   `public` `class` `Main {` `    ``public` `static` `void` `main(String[] args) {` `        ``/*` `        ``Input Tree:` `                ``5` `            ``/ \` `            ``4  6` `            ``\   \` `            ``3   7` `                    ``\` `                    ``8` `        ``*/` `        ``TreeNode root = ``new` `TreeNode(``5``);` `        ``root.left = ``new` `TreeNode(``4``);` `        ``root.left.right = ``new` `TreeNode(``3``);` `        ``root.right = ``new` `TreeNode(``6``);` `        ``root.right.right = ``new` `TreeNode(``7``);` `        ``root.right.right.right = ``new` `TreeNode(``8``);`   `        ``TreeNode p = root.left;` `        ``TreeNode q = root.left.right;`   `        ``Solution s = ``new` `Solution();` `        ``TreeNode lca1 = s.lowestCommonAncestor(root, p, q);` `        ``System.out.println(``"LCA of "` `+ p.val + ``" and "` `+ q.val + ``" is "` `+ lca1.val);`   `        ``TreeNode x = root.right.right;` `        ``TreeNode y = root.right.right.right;`   `        ``TreeNode lca2 = s.lowestCommonAncestor(root, x, y);` `        ``System.out.println(``"LCA of "` `+ x.val + ``" and "` `+ y.val + ``" is "` `+ lca2.val);` `    ``}` `}`

## Python3

 `class` `TreeNode:` `    ``def` `__init__(``self``, x):` `        ``self``.val ``=` `x` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None`   `# Function to find the lowest common ancestor (LCA)` `# of two nodes in a binary tree.` `def` `lowestCommonAncestor(root, p, q):` `    ``curr ``=` `root` `    ``while` `curr:` `        ``# If there is no left child, ` `        ``# traverse to the right child.` `        ``if` `curr.left ``is` `None``:` `            ``# Check if the current node is one of the target nodes.` `            ``if` `curr ``=``=` `p ``or` `curr ``=``=` `q:` `                ``return` `curr` `            ``curr ``=` `curr.right` `        ``else``:` `            ``# If there is a left child, find the in-order predecessor.` `            ``pre ``=` `curr.left` `            ``while` `pre.right ``and` `pre.right !``=` `curr:` `                ``pre ``=` `pre.right` `            ``if` `pre.right ``is` `None``:` `                ``# Set the right child of the ` `                ``# in-order predecessor to the current node.` `                ``pre.right ``=` `curr` `                ``curr ``=` `curr.left` `            ``else``:` `                ``# Remove the link from the in-order` `                ``# predecessor to the current node.` `                ``pre.right ``=` `None` `                ``# Check if the current node` `                ``# is one of the target nodes.` `                ``if` `curr ``=``=` `p ``or` `curr ``=``=` `q:` `                    ``return` `curr` `                ``curr ``=` `curr.right` `    ``return` `None`   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` `    ``"""` `    ``Input Tree:` `              ``5` `           ``/    \` `         ``4       6` `          ``\       \` `           ``3       7` `                    ``\` `                     ``8` `    ``"""` `    ``root ``=` `TreeNode(``5``)` `    ``root.left ``=` `TreeNode(``4``)` `    ``root.left.right ``=` `TreeNode(``3``)` `    ``root.right ``=` `TreeNode(``6``)` `    ``root.right.right ``=` `TreeNode(``7``)` `    ``root.right.right.right ``=` `TreeNode(``8``)`   `    ``p ``=` `root.left` `    ``q ``=` `root.left.right`   `    ``lca1 ``=` `lowestCommonAncestor(root, p, q)` `    ``print``(f``"LCA of {p.val} and {q.val} is {lca1.val}"``)`   `    ``x ``=` `root.right.right` `    ``y ``=` `root.right.right.right`   `    ``lca2 ``=` `lowestCommonAncestor(root, x, y)` `    ``print``(f``"LCA of {x.val} and {y.val} is {lca2.val}"``)`

## C#

 `using` `System;`   `public` `class` `TreeNode` `{` `    ``public` `int` `val;` `    ``public` `TreeNode left;` `    ``public` `TreeNode right;`   `    ``public` `TreeNode(``int` `x)` `    ``{` `        ``val = x;` `        ``left = ``null``;` `        ``right = ``null``;` `    ``}` `}`   `public` `class` `Solution` `{` `    ``// Function to find the lowest common ancestor (LCA)` `    ``// of two nodes in a binary tree.` `    ``public` `TreeNode LowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q)` `    ``{` `        ``TreeNode curr = root;` `        ``while` `(curr != ``null``)` `        ``{` `            ``// If there is no left child, ` `            ``// traverse to the right child.` `            ``if` `(curr.left == ``null``)` `            ``{` `                ``// Check if the current node is one of the target nodes.` `                ``if` `(curr == p || curr == q)` `                    ``return` `curr;` `                ``curr = curr.right;` `            ``}` `            ``else` `            ``{` `                ``// If there is a left child, ` `                ``// find the in-order predecessor.` `                ``TreeNode pre = curr.left;` `                ``while` `(pre.right != ``null` `&& pre.right != curr)` `                    ``pre = pre.right;` `                ``if` `(pre.right == ``null``)` `                ``{` `                    ``// Set the right child of the in-order` `                    ``// predecessor to the current node.` `                    ``pre.right = curr;` `                    ``curr = curr.left;` `                ``}` `                ``else` `                ``{` `                    ``// Remove the link from the in-order ` `                    ``// predecessor to the current node.` `                    ``pre.right = ``null``;` `                    ``// Check if the current node ` `                    ``// is one of the target nodes.` `                    ``if` `(curr == p || curr == q)` `                        ``return` `curr;` `                    ``curr = curr.right;` `                ``}` `            ``}` `        ``}` `        ``return` `null``;` `    ``}` `}`   `// Driver code` `public` `class` `GFG` `{` `    ``public` `static` `void` `Main(``string``[] args)` `    ``{` `        ``/*` `        ``Input Tree:` `                ``5` `            ``/ \` `            ``4  6` `            ``\   \` `             ``3   7` `                      ``\` `                       ``8` `        ``*/` `        ``TreeNode root = ``new` `TreeNode(5);` `        ``root.left = ``new` `TreeNode(4);` `        ``root.left.right = ``new` `TreeNode(3);` `        ``root.right = ``new` `TreeNode(6);` `        ``root.right.right = ``new` `TreeNode(7);` `        ``root.right.right.right = ``new` `TreeNode(8);`   `        ``TreeNode p = root.left;` `        ``TreeNode q = root.left.right;`   `        ``Solution s = ``new` `Solution();` `        ``TreeNode lca1 = s.LowestCommonAncestor(root, p, q);` `        ``Console.WriteLine(``"LCA of "` `+ p.val + ``" and "` `+ q.val + ``" is "` `+ lca1.val);`   `        ``TreeNode x = root.right.right;` `        ``TreeNode y = root.right.right.right;`   `        ``TreeNode lca2 = s.LowestCommonAncestor(root, x, y);` `        ``Console.WriteLine(``"LCA of "` `+ x.val + ``" and "` `+ y.val + ``" is "` `+ lca2.val);` `    ``}` `}`

## Javascript

 `class TreeNode {` `    ``constructor(x) {` `        ``this``.val = x;` `        ``this``.left = ``null``;` `        ``this``.right = ``null``;` `    ``}` `}`   `// Function to find the lowest common ancestor (LCA)` `// of two nodes in a binary tree.` `function` `lowestCommonAncestor(root, p, q) {` `    ``let curr = root;` `    ``while` `(curr !== ``null``) {` `        ``// If there is no left child, ` `        ``// traverse to the right child.` `        ``if` `(curr.left === ``null``) {` `            ``// Check if the current node is one of the target nodes.` `            ``if` `(curr === p || curr === q) ``return` `curr;` `            ``curr = curr.right;` `        ``} ``else` `{` `            ``// If there is a left child, ` `            ``// find the in-order predecessor.` `            ``let pre = curr.left;` `            ``while` `(pre.right !== ``null` `&& pre.right !== curr) pre = pre.right;` `            ``if` `(pre.right === ``null``) {` `                ``// Set the right child of the in-order ` `                ``// predecessor to the current node.` `                ``pre.right = curr;` `                ``curr = curr.left;` `            ``} ``else` `{` `                ``// Remove the link from the in-order ` `                ``// predecessor to the current node.` `                ``pre.right = ``null``;` `                ``// Check if the current node is one of the target nodes.` `                ``if` `(curr === p || curr === q) ``return` `curr;` `                ``curr = curr.right;` `            ``}` `        ``}` `    ``}` `    ``return` `null``;` `}`   `// Driver Code` `/*` `Input Tree:` `          ``5` `       ``/    \` `     ``4       6` `      ``\       \` `       ``3       7` `                ``\` `                 ``8` `*/` `const root = ``new` `TreeNode(5);` `root.left = ``new` `TreeNode(4);` `root.left.right = ``new` `TreeNode(3);` `root.right = ``new` `TreeNode(6);` `root.right.right = ``new` `TreeNode(7);` `root.right.right.right = ``new` `TreeNode(8);`   `const p = root.left;` `const q = root.left.right;`   `const lca1 = lowestCommonAncestor(root, p, q);` `console.log(`LCA of \${p.val} and \${q.val} is \${lca1.val}`);`   `const x = root.right.right;` `const y = root.right.right.right;`   `const lca2 = lowestCommonAncestor(root, x, y);` `console.log(`LCA of \${x.val} and \${y.val} is \${lca2.val}`);`

Output

``` LCA of 4 and 3 is 4
LCA of 7 and 8 is 7

```

Time Complexity: O(N) , The time complexity of the Morris Traversal approach to find the lowest common ancestor of two nodes in a binary search tree is O(N), where N is the number of nodes in the tree.

Auxiliary Space: O(1) , The space complexity of the Morris Traversal approach is O(1), which is constant extra space.