Lowest Common Ancestor in a Binary Tree | Set 1 - GeeksforGeeks

17 Apr, 2022
Given a binary tree (not a binary search tree) and two values say n1 and n2, write a program to find the least common ancestor.
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) 
 
We have discussed an efficient solution to find LCA in Binary Search Tree. In Binary Search Tree, using BST properties, we can find LCA in O(h) time where h is the height of the tree. Such an implementation is not possible in Binary Tree as keys Binary Tree nodes don’t follow any order. The following are different approaches to find LCA in Binary Tree.
Method 1 (By Storing root to n1 and root to n2 paths): Following is a simple O(n) algorithm to find LCA of n1 and n2. 1) Find a path from the root to n1 and store it in a vector or array. 2) Find a path from the root to n2 and store it in another vector or array. 3) Traverse both paths till the values in arrays are the same. Return the common element just before the mismatch. 
Following is the implementation of the above algorithm.
C++
Java
Python3
C#
Javascript
// C++ Program for Lowest Common Ancestor in a Binary Tree// A O(n) solution to find LCA of two given values n1 and n2#include <iostream>#include <vector> using namespace std; // A Binary Tree nodestruct Node{    int key;    struct Node *left, *right;}; // Utility function creates a new binary tree node with given keyNode * newNode(int k){    Node *temp = new Node;    temp->key = k;    temp->left = temp->right = NULL;    return temp;} // Finds the path from root node to given root of the tree, Stores the// path in a vector path[], returns true if path exists otherwise falsebool findPath(Node *root, vector<int> &path, int k){    // base case    if (root == NULL) return false;     // Store this node in path vector. The node will be removed if    // not in path from root to k    path.push_back(root->key);     // See if the k is same as root's key    if (root->key == k)        return true;     // Check if k is found in left or right sub-tree    if ( (root->left && findPath(root->left, path, k)) ||         (root->right && findPath(root->right, path, k)) )        return true;     // If not present in subtree rooted with root, remove root from    // path[] and return false    path.pop_back();    return false;} // Returns LCA if node n1, n2 are present in the given binary tree,// otherwise return -1int findLCA(Node *root, int n1, int n2){    // to store paths to n1 and n2 from the root    vector<int> path1, path2;     // Find paths from root to n1 and root to n1. If either n1 or n2    // is not present, return -1    if ( !findPath(root, path1, n1) || !findPath(root, path2, n2))          return -1;     /* Compare the paths to get the first different value */    int i;    for (i = 0; i < path1.size() && i < path2.size() ; i++)        if (path1[i] != path2[i])            break;    return path1[i-1];} // Driver program to test above functionsint main(){    // Let us create the Binary Tree shown in above diagram.    Node * root = newNode(1);    root->left = newNode(2);    root->right = newNode(3);    root->left->left = newNode(4);    root->left->right = newNode(5);    root->right->left = newNode(6);    root->right->right = newNode(7);    cout << "LCA(4, 5) = " << findLCA(root, 4, 5);    cout << "\nLCA(4, 6) = " << findLCA(root, 4, 6);    cout << "\nLCA(3, 4) = " << findLCA(root, 3, 4);    cout << "\nLCA(2, 4) = " << findLCA(root, 2, 4);    return 0;}
// Java Program for Lowest Common Ancestor in a Binary Tree// A O(n) solution to find LCA of two given values n1 and n2import java.util.ArrayList;import java.util.List; // A Binary Tree nodeclass Node {    int data;    Node left, right;     Node(int value) {        data = value;        left = right = null;    }} public class BT_NoParentPtr_Solution1{     Node root;    private List<Integer> path1 = new ArrayList<>();    private List<Integer> path2 = new ArrayList<>();     // Finds the path from root node to given root of the tree.    int findLCA(int n1, int n2) {        path1.clear();        path2.clear();        return findLCAInternal(root, n1, n2);    }     private int findLCAInternal(Node root, int n1, int n2) {         if (!findPath(root, n1, path1) || !findPath(root, n2, path2)) {            System.out.println((path1.size() > 0) ? "n1 is present" : "n1 is missing");            System.out.println((path2.size() > 0) ? "n2 is present" : "n2 is missing");            return -1;        }         int i;        for (i = 0; i < path1.size() && i < path2.size(); i++) {                     // System.out.println(path1.get(i) + " " + path2.get(i));            if (!path1.get(i).equals(path2.get(i)))                break;        }         return path1.get(i-1);    }         // Finds the path from root node to given root of the tree, Stores the    // path in a vector path[], returns true if path exists otherwise false    private boolean findPath(Node root, int n, List<Integer> path)    {        // base case        if (root == null) {            return false;        }                 // Store this node . The node will be removed if        // not in path from root to n.        path.add(root.data);         if (root.data == n) {            return true;        }         if (root.left != null && findPath(root.left, n, path)) {            return true;        }         if (root.right != null && findPath(root.right, n, path)) {            return true;        }         // If not present in subtree rooted with root, remove root from        // path[] and return false        path.remove(path.size()-1);         return false;    }     // Driver code    public static void main(String[] args)    {        BT_NoParentPtr_Solution1 tree = new BT_NoParentPtr_Solution1();        tree.root = new Node(1);        tree.root.left = new Node(2);        tree.root.right = new Node(3);        tree.root.left.left = new Node(4);        tree.root.left.right = new Node(5);        tree.root.right.left = new Node(6);        tree.root.right.right = new Node(7);         System.out.println("LCA(4, 5): " + tree.findLCA(4,5));        System.out.println("LCA(4, 6): " + tree.findLCA(4,6));        System.out.println("LCA(3, 4): " + tree.findLCA(3,4));        System.out.println("LCA(2, 4): " + tree.findLCA(2,4));         }}// This code is contributed by Sreenivasulu Rayanki.
# Python Program for Lowest Common Ancestor in a Binary Tree# O(n) solution to find LCS of two given values n1 and n2 # A binary tree nodeclass Node:    # Constructor to create a new binary node    def __init__(self, key):        self.key =  key        self.left = None        self.right = None # Finds the path from root node to given root of the tree.# Stores the path in a list path[], returns true if path# exists otherwise falsedef findPath( root, path, k):     # Baes Case    if root is None:        return False     # Store this node is path vector. The node will be    # removed if not in path from root to k    path.append(root.key)     # See if the k is same as root's key    if root.key == k :        return True     # Check if k is found in left or right sub-tree    if ((root.left != None and findPath(root.left, path, k)) or            (root.right!= None and findPath(root.right, path, k))):        return True     # If not present in subtree rooted with root, remove    # root from path and return False          path.pop()    return False # Returns LCA if node n1 , n2 are present in the given# binary tree otherwise return -1def findLCA(root, n1, n2):     # To store paths to n1 and n2 fromthe root    path1 = []    path2 = []     # Find paths from root to n1 and root to n2.    # If either n1 or n2 is not present , return -1    if (not findPath(root, path1, n1) or not findPath(root, path2, n2)):        return -1     # Compare the paths to get the first different value    i = 0    while(i < len(path1) and i < len(path2)):        if path1[i] != path2[i]:            break        i += 1    return path1[i-1]  # Driver program to test above function# Let's create the Binary Tree shown in above diagramroot = Node(1)root.left = Node(2)root.right = Node(3)root.left.left = Node(4)root.left.right = Node(5)root.right.left = Node(6)root.right.right = Node(7) print ("LCA(4, 5) = %d" %(findLCA(root, 4, 5,)))print ("LCA(4, 6) = %d" %(findLCA(root, 4, 6)))print ("LCA(3, 4) = %d" %(findLCA(root,3,4)))print ("LCA(2, 4) = %d" %(findLCA(root,2, 4))) # This code is contributed by Nikhil Kumar Singh(nickzuck_007)
// C# Program for Lowest Common// Ancestor in a Binary Tree// A O(n) solution to find LCA// of two given values n1 and n2using System.Collections;using System; // A Binary Tree nodeclass Node{  public int data;  public Node left, right;   public Node(int value)  {    data = value;    left = right = null;  }} public class BT_NoParentPtr_Solution1{   Node root;  private ArrayList path1 =          new ArrayList();  private ArrayList path2 =          new ArrayList();   // Finds the path from root  // node to given root of the  // tree.  int findLCA(int n1,              int n2)  {    path1.Clear();    path2.Clear();    return findLCAInternal(root,                           n1, n2);  } private int findLCAInternal(Node root,                            int n1, int n2){  if (!findPath(root, n1, path1) ||      !findPath(root, n2, path2)) {    Console.Write((path1.Count > 0) ?                  "n1 is present" :                  "n1 is missing");    Console.Write((path2.Count > 0) ?                  "n2 is present" :                  "n2 is missing");    return -1;  }   int i;  for (i = 0; i < path1.Count &&       i < path2.Count; i++)  {    // System.out.println(path1.get(i)    // + " " + path2.get(i));    if ((int)path1[i] !=        (int)path2[i])      break;  }  return (int)path1[i - 1];} // Finds the path from root node// to given root of the tree,// Stores the path in a vector// path[], returns true if path// exists otherwise falseprivate bool findPath(Node root,                      int n,                      ArrayList path){  // base case  if (root == null)  {    return false;  }   // Store this node . The node  // will be removed if not in  // path from root to n.  path.Add(root.data);   if (root.data == n)  {    return true;  }   if (root.left != null &&      findPath(root.left,               n, path))  {    return true;  }   if (root.right != null &&      findPath(root.right,               n, path))  {    return true;  }   // If not present in subtree  //rooted with root, remove root  // from path[] and return false  path.RemoveAt(path.Count - 1);   return false;} // Driver codepublic static void Main(String[] args){  BT_NoParentPtr_Solution1 tree =     new BT_NoParentPtr_Solution1();     tree.root = new Node(1);  tree.root.left = new Node(2);  tree.root.right = new Node(3);  tree.root.left.left = new Node(4);  tree.root.left.right = new Node(5);  tree.root.right.left = new Node(6);  tree.root.right.right = new Node(7);   Console.Write("LCA(4, 5): " +                 tree.findLCA(4, 5));  Console.Write("\nLCA(4, 6): " +                  tree.findLCA(4, 6));  Console.Write("\nLCA(3, 4): " +                  tree.findLCA(3, 4));  Console.Write("\nLCA(2, 4): " +                  tree.findLCA(2, 4));}} // This code is contributed by Rutvik_56
<script>     // JavaScript Program for Lowest Common    // Ancestor in a Binary Tree    // A O(n) solution to find LCA of    // two given values n1 and n2         class Node    {        constructor(value) {           this.left = null;           this.right = null;           this.data = value;        }    }         let root;    let path1 = [];    let path2 = [];      // Finds the path from root node to given root of the tree.    function findLCA(n1, n2) {        path1 = [];        path2 = [];        return findLCAInternal(root, n1, n2);    }      function findLCAInternal(root, n1, n2) {          if (!findPath(root, n1, path1) || !findPath(root, n2, path2))        {            document.write((path1.length > 0) ?            "n1 is present" : "n1 is missing");            document.write((path2.length > 0) ?            "n2 is present" : "n2 is missing");            return -1;        }          let i;        for (i = 0; i < path1.length && i < path2.length; i++) {                      // System.out.println(path1.get(i) + " " + path2.get(i));            if (path1[i] != path2[i])                break;        }          return path1[i-1];    }          // Finds the path from root node to    // given root of the tree, Stores the    // path in a vector path[], returns true    // if path exists otherwise false    function findPath(root, n, path)    {        // base case        if (root == null) {            return false;        }                  // Store this node . The node will be removed if        // not in path from root to n.        path.push(root.data);          if (root.data == n) {            return true;        }          if (root.left != null && findPath(root.left, n, path)) {            return true;        }          if (root.right != null && findPath(root.right, n, path)) {            return true;        }          // If not present in subtree rooted with root,        // remove root from        // path[] and return false        path.pop();          return false;    }         root = new Node(1);    root.left = new Node(2);    root.right = new Node(3);    root.left.left = new Node(4);    root.left.right = new Node(5);    root.right.left = new Node(6);    root.right.right = new Node(7);     document.write("LCA(4, 5) = " + findLCA(4,5) + "</br>");    document.write("LCA(4, 6) = " + findLCA(4,6) + "</br>");    document.write("LCA(3, 4) = " + findLCA(3,4) + "</br>");    document.write("LCA(2, 4) = " + findLCA(2,4)); </script>
Output: 
LCA(4, 5) = 2
LCA(4, 6) = 1
LCA(3, 4) = 1
LCA(2, 4) = 2
Time Complexity: The time complexity of the above solution is O(n). The tree is traversed twice, and then path arrays are compared. Thanks to Ravi Chandra Enaganti for suggesting the initial solution based on this method. 
Method 2 (Using Single Traversal) Method 1 finds LCA in O(n) time but requires three tree traversals plus extra spaces for path arrays. If we assume that the keys n1 and n2 are present in Binary Tree, we can find LCA using a single traversal of Binary Tree and without extra storage for path arrays. The idea is to traverse the tree starting from the root. If any of the given keys (n1 and n2) matches with the root, then the root is LCA (assuming that both keys are present). If the root doesn’t match with any of the keys, we recur for the left and right subtree. The node which has one key present in its left subtree and the other key present in the right subtree is the LCA. If both keys lie in the left subtree, then the left subtree has LCA also, otherwise, LCA lies in the right subtree.  
C++
C
Java
Python3
C#
Javascript
/* C++ Program to find LCA of n1 and n2 using one traversal of Binary Tree */#include <bits/stdc++.h>using namespace std; // A Binary Tree Nodestruct Node{    struct Node *left, *right;    int key;}; // Utility function to create a new tree NodeNode* newNode(int key){    Node *temp = new Node;    temp->key = key;    temp->left = temp->right = NULL;    return temp;} // This function returns pointer to LCA of two given values n1 and n2.// This function assumes that n1 and n2 are present in Binary Treestruct Node *findLCA(struct Node* root, int n1, int n2){    // Base case    if (root == NULL) return NULL;     // If either n1 or n2 matches with root's key, report    // the presence by returning root (Note that if a key is    // ancestor of other, then the ancestor key becomes LCA    if (root->key == n1 || root->key == n2)        return root;     // Look for keys in left and right subtrees    Node *left_lca  = findLCA(root->left, n1, n2);    Node *right_lca = findLCA(root->right, n1, n2);     // If both of the above calls return Non-NULL, then one key    // is present in once subtree and other is present in other,    // So this node is the LCA    if (left_lca && right_lca)  return root;     // Otherwise check if left subtree or right subtree is LCA    return (left_lca != NULL)? left_lca: right_lca;} // Driver program to test above functionsint main(){    // Let us create binary tree given in the above example    Node * root = newNode(1);    root->left = newNode(2);    root->right = newNode(3);    root->left->left = newNode(4);    root->left->right = newNode(5);    root->right->left = newNode(6);    root->right->right = newNode(7);    cout << "LCA(4, 5) = " << findLCA(root, 4, 5)->key;    cout << "\nLCA(4, 6) = " << findLCA(root, 4, 6)->key;    cout << "\nLCA(3, 4) = " << findLCA(root, 3, 4)->key;    cout << "\nLCA(2, 4) = " << findLCA(root, 2, 4)->key;    return 0;} // This code is contributed by Aditya Kumar (adityakumar129)
// C Program to find LCA of n1 and n2 using one traversalof// Binary Tree#include <stdio.h>#include <stdlib.h> // A Binary Tree Nodetypedef struct Node {    struct Node *left, *right;    int key;}Node; // Utility function to create a new tree NodeNode* newNode(int key){    Node* temp = (Node*)malloc(sizeof(Node));    temp->key = key;    temp->left = temp->right = NULL;    return temp;} // This function returns pointer to LCA of two given values// n1 and n2. This function assumes that n1 and n2 are// present in Binary TreeNode* findLCA(Node* root, int n1, int n2){    // Base case    if (root == NULL)        return NULL;     // If either n1 or n2 matches with root's key, report    // the presence by returning root (Note that if a key is    // ancestor of other, then the ancestor key becomes LCA    if (root->key == n1 || root->key == n2)        return root;     // Look for keys in left and right subtrees    Node* left_lca = findLCA(root->left, n1, n2);    Node* right_lca = findLCA(root->right, n1, n2);     // If both of the above calls return Non-NULL, then one    // key is present in once subtree and other is present    // in other, So this node is the LCA    if (left_lca && right_lca)        return root;     // Otherwise check if left subtree or right subtree is    // LCA    return (left_lca != NULL) ? left_lca : right_lca;} // Driver program to test above functionsint main(){    // Let us create binary tree given in the above example    Node* root = newNode(1);    root->left = newNode(2);    root->right = newNode(3);    root->left->left = newNode(4);    root->left->right = newNode(5);    root->right->left = newNode(6);    root->right->right = newNode(7);    printf("LCA(4, 5) = %d", findLCA(root, 4, 5)->key);    printf("\nLCA(4, 6) = %d", findLCA(root, 4, 6)->key);    printf("\nLCA(3, 4) = %d", findLCA(root, 3, 4)->key);    printf("\nLCA(2, 4) = %d", findLCA(root, 2, 4)->key);    return 0;} // This code is contributed by Aditya Kumar (adityakumar129)
//Java implementation to find lowest common ancestor of// n1 and n2 using one traversal of binary tree /* Class containing left and right child of current node and key value*/class Node{    int data;    Node left, right;     public Node(int item)    {        data = item;        left = right = null;    }} public class BinaryTree{    //Root of the Binary Tree    Node root;     Node findLCA(int n1, int n2)    {        return findLCA(root, n1, n2);    }     // This function returns pointer to LCA of two given    // values n1 and n2. This function assumes that n1 and    // n2 are present in Binary Tree    Node findLCA(Node node, int n1, int n2)    {        // Base case        if (node == null)            return null;         // If either n1 or n2 matches with root's key, report        // the presence by returning root (Note that if a key is        // ancestor of other, then the ancestor key becomes LCA        if (node.data == n1 || node.data == n2)            return node;         // Look for keys in left and right subtrees        Node left_lca = findLCA(node.left, n1, n2);        Node right_lca = findLCA(node.right, n1, n2);         // If both of the above calls return Non-NULL, then one key        // is present in once subtree and other is present in other,        // So this node is the LCA        if (left_lca!=null && right_lca!=null)            return node;         // Otherwise check if left subtree or right subtree is LCA        return (left_lca != null) ? left_lca : right_lca;    }     /* Driver program to test above functions */    public static void main(String args[])    {        BinaryTree tree = new BinaryTree();        tree.root = new Node(1);        tree.root.left = new Node(2);        tree.root.right = new Node(3);        tree.root.left.left = new Node(4);        tree.root.left.right = new Node(5);        tree.root.right.left = new Node(6);        tree.root.right.right = new Node(7);        System.out.println("LCA(4, 5) = " +                            tree.findLCA(4, 5).data);        System.out.println("LCA(4, 6) = " +                            tree.findLCA(4, 6).data);        System.out.println("LCA(3, 4) = " +                            tree.findLCA(3, 4).data);        System.out.println("LCA(2, 4) = " +                            tree.findLCA(2, 4).data);    }}
# Python program to find LCA of n1 and n2 using one# traversal of Binary tree # A binary tree nodeclass Node:         # Constructor to create a new tree node    def __init__(self, key):        self.key = key        self.left = None        self.right = None     # This function returns pointer to LCA of two given# values n1 and n2# This function assumes that n1 and n2 are present in# Binary Treedef findLCA(root, n1, n2):         # Base Case    if root is None:        return None     # If either n1 or n2 matches with root's key, report    #  the presence by returning root (Note that if a key is    #  ancestor of other, then the ancestor key becomes LCA    if root.key == n1 or root.key == n2:        return root     # Look for keys in left and right subtrees    left_lca = findLCA(root.left, n1, n2)    right_lca = findLCA(root.right, n1, n2)     # If both of the above calls return Non-NULL, then one key    # is present in once subtree and other is present in other,    # So this node is the LCA    if left_lca and right_lca:        return root     # Otherwise check if left subtree or right subtree is LCA    return left_lca if left_lca is not None else right_lca  # Driver program to test above function # Let us create a binary tree given in the above exampleroot = Node(1)root.left = Node(2)root.right = Node(3)root.left.left = Node(4)root.left.right = Node(5)root.right.left = Node(6)root.right.right = Node(7)print ("LCA(4,5) = ", findLCA(root, 4, 5).key)print ("LCA(4,6) = ", findLCA(root, 4, 6).key)print ("LCA(3,4) = ", findLCA(root, 3, 4).key)print ("LCA(2,4) = ", findLCA(root, 2, 4).key) # This code is contributed by Nikhil Kumar Singh(nickzuck_007)
// C# implementation to find lowest common// ancestor of n1 and n2 using one traversal// of binary treeusing System; // Class containing left and right// child of current node and key valuepublic class Node{    public int data;    public Node left, right;         public Node(int item)    {        data = item;        left = right = null;    }} class BinaryTree{     // Root of the Binary TreeNode root; Node findLCA(int n1, int n2){    return findLCA(root, n1, n2);} // This function returns pointer to LCA// of two given values n1 and n2. This// function assumes that n1 and n2 are// present in Binary TreeNode findLCA(Node node, int n1, int n2){         // Base case    if (node == null)        return null;             // If either n1 or n2 matches with    // root's key, report the presence    // by returning root (Note that if    // a key is ancestor of other,    // then the ancestor key becomes LCA    if (node.data == n1 || node.data == n2)        return node;     // Look for keys in left and right subtrees    Node left_lca = findLCA(node.left, n1, n2);    Node right_lca = findLCA(node.right, n1, n2);     // If both of the above calls return Non-NULL,    // then one key is present in once subtree    // and other is present in other, So this    // node is the LCA    if (left_lca != null && right_lca != null)        return node;     // Otherwise check if left subtree or    // right subtree is LCA    return (left_lca != null) ? left_lca : right_lca;} // Driver codepublic static void Main(string []args){    BinaryTree tree = new BinaryTree();    tree.root = new Node(1);    tree.root.left = new Node(2);    tree.root.right = new Node(3);    tree.root.left.left = new Node(4);    tree.root.left.right = new Node(5);    tree.root.right.left = new Node(6);    tree.root.right.right = new Node(7);         Console.WriteLine("LCA(4, 5) = " +                      tree.findLCA(4, 5).data);    Console.WriteLine("LCA(4, 6) = " +                      tree.findLCA(4, 6).data);    Console.WriteLine("LCA(3, 4) = " +                      tree.findLCA(3, 4).data);    Console.WriteLine("LCA(2, 4) = " +                      tree.findLCA(2, 4).data);}} // This code is contributed by pratham76
<script>     // JavaScript implementation to find    // lowest common ancestor of    // n1 and n2 using one traversal of binary tree         class Node    {        constructor(item) {           this.left = null;           this.right = null;           this.data = item;        }    }         //Root of the Binary Tree    let root;      function findlCA(n1, n2)    {        return findLCA(root, n1, n2);    }      // This function returns pointer to LCA of two given    // values n1 and n2. This function assumes that n1 and    // n2 are present in Binary Tree    function findLCA(node, n1, n2)    {        // Base case        if (node == null)            return null;          // If either n1 or n2 matches with root's key, report        // the presence by returning root (Note that if a key is        // ancestor of other, then the ancestor key becomes LCA        if (node.data == n1 || node.data == n2)            return node;          // Look for keys in left and right subtrees        let left_lca = findLCA(node.left, n1, n2);        let right_lca = findLCA(node.right, n1, n2);          // If both of the above calls return Non-NULL, then one key        // is present in once subtree and other is present in other,        // So this node is the LCA        if (left_lca!=null && right_lca!=null)            return node;          // Otherwise check if left subtree or right subtree is LCA        return (left_lca != null) ? left_lca : right_lca;    }         root = new Node(1);    root.left = new Node(2);    root.right = new Node(3);    root.left.left = new Node(4);    root.left.right = new Node(5);    root.right.left = new Node(6);    root.right.right = new Node(7);    document.write("LCA(4, 5) = " +                       findlCA(4, 5).data + "</br>");    document.write("LCA(4, 6) = " +                       findlCA(4, 6).data + "</br>");    document.write("LCA(3, 4) = " +                       findlCA(3, 4).data + "</br>");    document.write("LCA(2, 4) = " +                       findlCA(2, 4).data + "</br>");     </script>
Output: 
LCA(4, 5) = 2
LCA(4, 6) = 1
LCA(3, 4) = 1
LCA(2, 4) = 2
Thanks to Atul Singh for suggesting this solution. 
Time Complexity: The time complexity of the above solution is O(n) as the method does a simple tree traversal in a bottom-up fashion. 
Note that the above method assumes that keys are present in Binary Tree. If one key is present and the other is absent, then it returns the present key as LCA (Ideally should have returned NULL). 
We can extend this method to handle all cases bypassing two boolean variables v1 and v2. v1 is set as true when n1 is present in the tree and v2 is set as true if n2 is present in the tree. 
C++
Java
Python3
C#
Javascript
/* C++ program to find LCA of n1 and n2 using one traversal of Binary Tree.   It handles all cases even when n1 or n2 is not there in Binary Tree */#include <iostream>using namespace std; // A Binary Tree Nodestruct Node{    struct Node *left, *right;    int key;}; // Utility function to create a new tree NodeNode* newNode(int key){    Node *temp = new Node;    temp->key = key;    temp->left = temp->right = NULL;    return temp;} // This function returns pointer to LCA of two given values n1 and n2.// v1 is set as true by this function if n1 is found// v2 is set as true by this function if n2 is foundstruct Node *findLCAUtil(struct Node* root, int n1, int n2, bool &v1, bool &v2){    // Base case    if (root == NULL) return NULL;     // If either n1 or n2 matches with root's key, report the presence    // by setting v1 or v2 as true and return root (Note that if a key    // is ancestor of other, then the ancestor key becomes LCA)    if (root->key == n1)    {        v1 = true;        return root;    }    if (root->key == n2)    {        v2 = true;        return root;    }     // Look for keys in left and right subtrees    Node *left_lca  = findLCAUtil(root->left, n1, n2, v1, v2);    Node *right_lca = findLCAUtil(root->right, n1, n2, v1, v2);     // If both of the above calls return Non-NULL, then one key    // is present in once subtree and other is present in other,    // So this node is the LCA    if (left_lca && right_lca)  return root;     // Otherwise check if left subtree or right subtree is LCA    return (left_lca != NULL)? left_lca: right_lca;} // Returns true if key k is present in tree rooted with rootbool find(Node *root, int k){    // Base Case    if (root == NULL)        return false;     // If key is present at root, or in left subtree or right subtree,    // return true;    if (root->key == k || find(root->left, k) ||  find(root->right, k))        return true;     // Else return false    return false;} // This function returns LCA of n1 and n2 only if both n1 and n2 are present// in tree, otherwise returns NULL;Node *findLCA(Node *root, int n1, int n2){    // Initialize n1 and n2 as not visited    bool v1 = false, v2 = false;     // Find lca of n1 and n2 using the technique discussed above    Node *lca = findLCAUtil(root, n1, n2, v1, v2);     // Return LCA only if both n1 and n2 are present in tree    if (v1 && v2 || v1 && find(lca, n2) || v2 && find(lca, n1))        return lca;     // Else return NULL    return NULL;} // Driver program to test above functionsint main(){    // Let us create binary tree given in the above example    Node * root = newNode(1);    root->left = newNode(2);    root->right = newNode(3);    root->left->left = newNode(4);    root->left->right = newNode(5);    root->right->left = newNode(6);    root->right->right = newNode(7);    Node *lca =  findLCA(root, 4, 5);    if (lca != NULL)       cout << "LCA(4, 5) = " << lca->key;    else       cout << "Keys are not present ";     lca =  findLCA(root, 4, 10);    if (lca != NULL)       cout << "\nLCA(4, 10) = " << lca->key;    else       cout << "\nKeys are not present ";     return 0;}
// Java implementation to find lowest common ancestor of// n1 and n2 using one traversal of binary tree// It also handles cases even when n1 and n2 are not there in Tree /* Class containing left and right child of current node and key */class Node{    int data;    Node left, right;     public Node(int item)    {        data = item;        left = right = null;    }} public class BinaryTree{    // Root of the Binary Tree    Node root;    static boolean v1 = false, v2 = false;     // This function returns pointer to LCA of two given    // values n1 and n2.    // v1 is set as true by this function if n1 is found    // v2 is set as true by this function if n2 is found    Node findLCAUtil(Node node, int n1, int n2)    {        // Base case        if (node == null)            return null;                 //Store result in temp, in case of key match so that we can search for other key also.        Node temp=null;         // If either n1 or n2 matches with root's key, report the presence        // by setting v1 or v2 as true and return root (Note that if a key        // is ancestor of other, then the ancestor key becomes LCA)        if (node.data == n1)        {            v1 = true;            temp = node;        }        if (node.data == n2)        {            v2 = true;            temp = node;        }         // Look for keys in left and right subtrees        Node left_lca = findLCAUtil(node.left, n1, n2);        Node right_lca = findLCAUtil(node.right, n1, n2);         if (temp != null)            return temp;         // If both of the above calls return Non-NULL, then one key        // is present in once subtree and other is present in other,        // So this node is the LCA        if (left_lca != null && right_lca != null)            return node;         // Otherwise check if left subtree or right subtree is LCA        return (left_lca != null) ? left_lca : right_lca;    }     // Finds lca of n1 and n2 under the subtree rooted with 'node'    Node findLCA(int n1, int n2)    {        // Initialize n1 and n2 as not visited        v1 = false;        v2 = false;         // Find lca of n1 and n2 using the technique discussed above        Node lca = findLCAUtil(root, n1, n2);         // Return LCA only if both n1 and n2 are present in tree        if (v1 && v2)            return lca;         // Else return NULL        return null;    }     /* Driver program to test above functions */    public static void main(String args[])    {        BinaryTree tree = new BinaryTree();        tree.root = new Node(1);        tree.root.left = new Node(2);        tree.root.right = new Node(3);        tree.root.left.left = new Node(4);        tree.root.left.right = new Node(5);        tree.root.right.left = new Node(6);        tree.root.right.right = new Node(7);         Node lca = tree.findLCA(4, 5);        if (lca != null)            System.out.println("LCA(4, 5) = " + lca.data);        else            System.out.println("Keys are not present");         lca = tree.findLCA(4, 10);        if (lca != null)            System.out.println("LCA(4, 10) = " + lca.data);        else            System.out.println("Keys are not present");    }}
""" Program to find LCA of n1 and n2 using one traversal of Binary treeIt handles all cases even when n1 or n2 is not there in tree""" # A binary tree nodeclass Node:     # Constructor to create a new node    def __init__(self, key):        self.key = key        self.left = None        self.right = None # This function return pointer to LCA of two given values# n1 and n2# v1 is set as true by this function if n1 is found# v2 is set as true by this function if n2 is founddef findLCAUtil(root, n1, n2, v):         # Base Case    if root is None:        return None     # IF either n1 or n2 matches ith root's key, report    # the presence by setting v1 or v2 as true and return    # root (Note that if a key is ancestor of other, then    # the ancestor key becomes LCA)    if root.key == n1 :        v[0] = True        return root     if root.key == n2:        v[1] = True        return root     # Look for keys in left and right subtree    left_lca = findLCAUtil(root.left, n1, n2, v)    right_lca = findLCAUtil(root.right, n1, n2, v)     # If both of the above calls return Non-NULL, then one key    # is present in once subtree and other is present in other,    # So this node is the LCA    if left_lca and right_lca:        return root     # Otherwise check if left subtree or right subtree is LCA    return left_lca if left_lca is not None else right_lca  def find(root, k):         # Base Case    if root is None:        return False         # If key is present at root, or if left subtree or right    # subtree , return true    if (root.key == k or find(root.left, k) or        find(root.right, k)):        return True         # Else return false    return False # This function returns LCA of n1 and n2 on value if both# n1 and n2 are present in tree, otherwise returns Nonedef findLCA(root, n1, n2):         # Initialize n1 and n2 as not visited    v = [False, False]     # Find lca of n1 and n2 using the technique discussed above    lca = findLCAUtil(root, n1, n2, v)     # Returns LCA only if both n1 and n2 are present in tree    if (v[0] and v[1] or v[0] and find(lca, n2) or v[1] and        find(lca, n1)):        return lca     # Else return None    return None # Driver program to test above functionroot = Node(1)root.left = Node(2)root.right = Node(3)root.left.left = Node(4)root.left.right = Node(5)root.right.left = Node(6)root.right.right = Node(7) lca = findLCA(root, 4, 5) if lca is not None:    print ("LCA(4, 5) = ", lca.key)else :    print ("Keys are not present") lca = findLCA(root, 4, 10)if lca is not None:    print ("LCA(4,10) = ", lca.key)else:    print ("Keys are not present") # This code is contributed by Nikhil Kumar Singh(nickzuck_007)
using System; // c# implementation to find lowest common ancestor of// n1 and n2 using one traversal of binary tree// It also handles cases even when n1 and n2 are not there in Tree /* Class containing left and right child of current node and key */public class Node{    public int data;    public Node left, right;     public Node(int item)    {        data = item;        left = right = null;    }} public class BinaryTree{    // Root of the Binary Tree    public Node root;    public static bool v1 = false, v2 = false;     // This function returns pointer to LCA of two given    // values n1 and n2.    // v1 is set as true by this function if n1 is found    // v2 is set as true by this function if n2 is found    public virtual Node findLCAUtil(Node node, int n1, int n2)    {        // Base case        if (node == null)        {            return null;        }         //Store result in temp, in case of key match so that we can search for other key also.        Node temp = null;         // If either n1 or n2 matches with root's key, report the presence        // by setting v1 or v2 as true and return root (Note that if a key        // is ancestor of other, then the ancestor key becomes LCA)        if (node.data == n1)        {            v1 = true;            temp = node;        }        if (node.data == n2)        {            v2 = true;            temp = node;        }         // Look for keys in left and right subtrees        Node left_lca = findLCAUtil(node.left, n1, n2);        Node right_lca = findLCAUtil(node.right, n1, n2);         if (temp != null)        {            return temp;        }         // If both of the above calls return Non-NULL, then one key        // is present in once subtree and other is present in other,        // So this node is the LCA        if (left_lca != null && right_lca != null)        {            return node;        }         // Otherwise check if left subtree or right subtree is LCA        return (left_lca != null) ? left_lca : right_lca;    }     // Finds lca of n1 and n2 under the subtree rooted with 'node'    public virtual Node findLCA(int n1, int n2)    {        // Initialize n1 and n2 as not visited        v1 = false;        v2 = false;         // Find lca of n1 and n2 using the technique discussed above        Node lca = findLCAUtil(root, n1, n2);         // Return LCA only if both n1 and n2 are present in tree        if (v1 && v2)        {            return lca;        }         // Else return NULL        return null;    }     /* Driver program to test above functions */    public static void Main(string[] args)    {        BinaryTree tree = new BinaryTree();        tree.root = new Node(1);        tree.root.left = new Node(2);        tree.root.right = new Node(3);        tree.root.left.left = new Node(4);        tree.root.left.right = new Node(5);        tree.root.right.left = new Node(6);        tree.root.right.right = new Node(7);         Node lca = tree.findLCA(4, 5);        if (lca != null)        {            Console.WriteLine("LCA(4, 5) = " + lca.data);        }        else        {            Console.WriteLine("Keys are not present");        }         lca = tree.findLCA(4, 10);        if (lca != null)        {            Console.WriteLine("LCA(4, 10) = " + lca.data);        }        else        {            Console.WriteLine("Keys are not present");        }    }}   // This code is contributed by Shrikant13
<script> // JavaScript implementation to find lowest// common ancestor of n1 and n2 using one// traversal of binary tree. It also handles// cases even when n1 and n2 are not there in Tree // Class containing left and right child// of current node and keyclass Node{    constructor(item)    {        this.data = item;        this.left = null;        this.right = null;    }} class BinaryTree{     // Root of the Binary Treeconstructor(){    this.root = null;    this.v1 = false;    this.v2 = false;} // This function returns pointer to LCA// of two given values n1 and n2.// v1 is set as true by this function// if n1 is found// v2 is set as true by this function// if n2 is foundfindLCAUtil(node, n1, n2){         // Base case    if (node == null)    {        return null;    }         // Store result in temp, in case of    // key match so that we can search    // for other key also.    var temp = null;         // If either n1 or n2 matches with root's key,    // report the presence by setting v1 or v2 as    // true and return root (Note that if a key    // is ancestor of other, then the ancestor    // key becomes LCA)    if (node.data == n1)    {        this.v1 = true;        temp = node;    }    if (node.data == n2)    {        this.v2 = true;        temp = node;    }         // Look for keys in left and right subtrees    var left_lca = this.findLCAUtil(node.left, n1, n2);    var right_lca = this.findLCAUtil(node.right, n1, n2);         if (temp != null)    {        return temp;    }         // If both of the above calls return Non-NULL,    // then one key is present in once subtree and    // other is present in other, So this node is the LCA    if (left_lca != null && right_lca != null)    {        return node;    }         // Otherwise check if left subtree or    // right subtree is LCA    return left_lca != null ? left_lca : right_lca;} // Finds lca of n1 and n2 under the// subtree rooted with 'node'findLCA(n1, n2){         // Initialize n1 and n2 as not visited    this.v1 = false;    this.v2 = false;         // Find lca of n1 and n2 using the    // technique discussed above    var lca = this.findLCAUtil(this.root, n1, n2);         // Return LCA only if both n1 and n2    // are present in tree    if (this.v1 && this.v2)    {        return lca;    }         // Else return NULL    return null;}} // Driver codevar tree = new BinaryTree();tree.root = new Node(1);tree.root.left = new Node(2);tree.root.right = new Node(3);tree.root.left.left = new Node(4);tree.root.left.right = new Node(5);tree.root.right.left = new Node(6);tree.root.right.right = new Node(7); var lca = tree.findLCA(4, 5);if (lca != null){    document.write("LCA(4, 5) = " +                   lca.data + "<br>");} else{    document.write("Keys are not present" + "<br>");} lca = tree.findLCA(4, 10);if (lca != null){    document.write("LCA(4, 10) = " +                   lca.data + "<br>");}else{    document.write("Keys are not present" + "<br>");} // This code is contributed by rdtank </script>
Output: 
LCA(4, 5) = 2
Keys are not present
Thanks to Dhruv for suggesting this extended solution. You may like to see the below articles as well : LCA using Parent Pointer 
Lowest Common Ancestor in a Binary Search Tree. 
Find LCA in Binary Tree using RMQPlease write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Tushar Garg 6
shrikanth13
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pratham76
mukesh07
rdtank
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Tree
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