# Lowest Common Ancestor in a Binary Search Tree.

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.

**Examples:**

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Tree:

Input:LCA of 10 and 14Output:12Explanation:12 is the closest node to both 10 and 14 which is a ancestor of both the nodes.Input:LCA of 8 and 14Output:8Explanation:8 is the closest node to both 8 and 14 which is a ancestor of both the nodes.Input:LCA of 10 and 22Output:20Explanation:20 is the closest node to both 10 and 22 which is a ancestor of both the nodes.

**Following is the** **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)

**Approach:** 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 in, if node’s value is greater than both n1 and n2 then our LCA lies in the left side of the node, if it’s 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).

**Algorithm:**

- 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.

**Implementation:**

## C++

`// A recursive CPP program to find` `// LCA of two nodes n1 and n2.` `#include <bits/stdc++.h>` `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);` ` ` `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 <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)` |

## C#

`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 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);` ` ` `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

`<script>` `// Recursive JavaScript program to print lca of two nodes` ` ` `// A binary tree node` `class Node` `{` ` ` `constructor(item)` ` ` `{` ` ` `this` `.data=item;` ` ` `this` `.left=` `this` `.right=` `null` `;` ` ` `}` `}` `let root;` `function` `lca(node,n1,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() */` ` ` `// Let us construct the BST shown in the above figure` `root = ` `new` `Node(20);` `root.left = ` `new` `Node(8);` `root.right = ` `new` `Node(22);` `root.left.left = ` `new` `Node(4);` `root.left.right = ` `new` `Node(12);` `root.left.right.left = ` `new` `Node(10);` `root.left.right.right = ` `new` `Node(14);` `let n1 = 10, n2 = 14;` `let t = lca(root, n1, n2);` `document.write(` `"LCA of "` `+ n1 + ` `" and "` `+ n2 + ` `" is "` `+` `t.data+` `"<br>"` `);` `n1 = 14;` `n2 = 8;` `t = lca(root, n1, n2);` `document.write(` `"LCA of "` `+ n1 + ` `" and "` `+ n2 + ` `" is "` `+` `t.data+` `"<br>"` `);` `n1 = 10;` `n2 = 22;` `t = lca(root, n1, n2);` `document.write(` `"LCA of "` `+ n1 + ` `" and "` `+ n2 + ` `" is "` `+` `t.data+` `"<br>"` `);` `// This code is contributed by avanitrachhadiya2155` `</script>` |

**Output**

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

**Complexity Analysis:**

**Time Complexity:**O(h).

The time Complexity of the above solution is O(h), where h is the height of the tree.**Space Complexity:**O(h).

If recursive stack space is ignored, the space complexity of the above solution is constant.

**Iterative Implementation:** The above solution uses recursion. The recursive solution requires extra space in the form of the function call stack. So an iterative solution can be implemented which does not occupy space in the form of the function call stack.

**Implementation:**

## C++

`// A recursive CPP program to find` `// LCA of two nodes n1 and n2.` `#include <bits/stdc++.h>` `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 */` `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.*/` `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);` ` ` `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 <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)` `{` ` ` `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 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 */` `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 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 is contributed by SHUBHAMSINGH10` |

## 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):` ` ` `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 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 Sumit Bhardwaj (Timus)` |

## C#

`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 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);` ` ` `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

`<script>` `// Recursive Javascript program to` `// print lca of two nodes` `// A binary tree node` `class Node` `{` ` ` `constructor(item)` ` ` `{` ` ` `this` `.data = item;` ` ` `this` `.left = ` `null` `;` ` ` `this` `.right = ` `null` `;` ` ` `}` `}` `var` `root = ` `null` `;` `/* Function to find LCA of n1 and n2.` `The function assumes that both` `n1 and n2 are present in BST */` `function` `lca(root, n1, 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` `// Let us construct the BST shown` `// in the above figure` `root = ` `new` `Node(20);` `root.left = ` `new` `Node(8);` `root.right = ` `new` `Node(22);` `root.left.left = ` `new` `Node(4);` `root.left.right = ` `new` `Node(12);` `root.left.right.left = ` `new` `Node(10);` `root.left.right.right = ` `new` `Node(14);` `var` `n1 = 10, n2 = 14;` `var` `t = lca(root, n1, n2);` `document.write(` `"LCA of "` `+ n1 + ` `" and "` `+ n2 +` ` ` `" is "` `+ t.data + ` `"<br>"` `);` `n1 = 14;` `n2 = 8;` `t = lca(root, n1, n2);` `document.write(` `"LCA of "` `+ n1 + ` `" and "` `+ n2 +` ` ` `" is "` `+ t.data+ ` `"<br>"` `);` `n1 = 10;` `n2 = 22;` `t = lca(root, n1, n2);` `document.write(` `"LCA of "` `+ n1 + ` `" and "` `+ n2 +` ` ` `" is "` `+ t.data+ ` `"<br>"` `);` ` ` `// This code is contributed by rrrtnx` `</script>` |

**Output**

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

**Complexity Analysis:**

**Time Complexity:**O(h).

The Time Complexity of the above solution is O(h), where h is the height of the tree.**Space Complexity:**O(1).

The space complexity of the above solution is constant.

https://www.youtube.com/watch?v=zlTsz

-apm4U

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 an incorrect result. Extend the above solutions to return NULL if n1 or n2 or both not present in BST.

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