Lowest Common Ancestor in a Binary Search Tree.

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

Examples:

Tree: 
BST_LCA

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.

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

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)



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:

  1. Create a recursive function that takes a node and the two values n1 and n2.
  2. 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 thr right subtree.
  3. 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 thr left subtree.
  4. If both the above cases are false then return the current node as LCA.

Implementation:

C

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// 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;
}

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C++

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// 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

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// 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

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Python

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# 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)

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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

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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).
    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

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// 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;
}

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// 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

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Java

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// 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

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# 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)

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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

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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.

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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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