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 <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); // 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 <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 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
<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> |
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 <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) { 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 <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 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
<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> |
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.
Related Articles: Lowest Common Ancestor in a Binary Tree, LCA using Parent Pointer, Find LCA in Binary Tree using RMQ
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Please Login to comment...