Lowest Common Ancestor in a Binary Search Tree.

Check if the given array can represent Level Order Traversal of Binary Search Tree

Find k-th smallest element in BST (Order Statistics in BST)

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

// 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

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


# 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)

// 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

// 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);
console.log("LCA of " + n1 + " and " + n2 + " is " +
t.data+"<br>");

n1 = 14;
n2 = 8;
t = lca(root, n1, n2);
console.log("LCA of " + n1 + " and " + n2 + " is " + 
t.data+"<br>");

n1 = 10;
n2 = 22;
t = lca(root, n1, n2);
console.log("LCA of " + n1 + " and " + n2 + " is " + 
t.data+"<br>");

// This code is contributed by avanitrachhadiya2155

Output

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

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

Below is the iterative implementation of the above approach:

C++

// A recursive CPP program to find
// LCA of two nodes n1 and n2.
#include <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

// 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

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 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)

// 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

// 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);
console.log("LCA of " + n1 + " and " + n2 + 
               " is " + t.data + "<br>");

n1 = 14;
n2 = 8;
t = lca(root, n1, n2);
console.log("LCA of " + n1 + " and " + n2 + 
               " is " + t.data+ "<br>");

n1 = 10;
n2 = 22;
t = lca(root, n1, n2);
console.log("LCA of " + n1 + " and " + n2 + 
               " is " + t.data+ "<br>");
               
// This code is contributed by rrrtnx

Output

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

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

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

Follow the steps to implement the above approach:

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

Below is the implementation of the above approach:

C++

#include <iostream>
using namespace std;

struct TreeNode {
    int val;
    TreeNode *left, *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
    if (root == NULL || root == p || root == q) {
        return root;
    }

    TreeNode* leftLCA = lowestCommonAncestor(root->left, p, q);
    TreeNode* rightLCA = lowestCommonAncestor(root->right, p, q);

    if (leftLCA != NULL && rightLCA != NULL) {
        return root;
    }

    return (leftLCA != NULL) ? leftLCA : rightLCA;
}

// Driver Code
int main() {
    /*
    Input Tree:
            5
        / \
        4   6
        \    \
        3    7
                \
                8
    */
    TreeNode* root = new TreeNode(5);
    root->left = new TreeNode(4);
    root->left->right = new TreeNode(3);
    root->right = new TreeNode(6);
    root->right->right = new TreeNode(7);
    root->right->right->right = new TreeNode(8);

    TreeNode* p = root->left;
    TreeNode* q = root->left->right;

    TreeNode* lca1 = lowestCommonAncestor(root, p, q);
    cout << "LCA of " << p->val << " and " << q->val << " is " << lca1->val << endl;

    TreeNode* x = root->right->right;
    TreeNode* y = root->right->right->right;

    TreeNode* lca2 = lowestCommonAncestor(root, x, y);
    cout << "LCA of " << x->val << " and " << y->val << " is " << lca2->val << endl;

    return 0;
}

Java

class TreeNode {
    int val;
    TreeNode left, right;
    
    public TreeNode(int x) {
        val = x;
        left = right = null;
    }
}

public class LowestCommonAncestor {
    public static TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null || root == p || root == q) {
            return root;
        }

        TreeNode leftLCA = lowestCommonAncestor(root.left, p, q);
        TreeNode rightLCA = lowestCommonAncestor(root.right, p, q);

        if (leftLCA != null && rightLCA != null) {
            return root;
        }

        return (leftLCA != null) ? leftLCA : rightLCA;
    }

    public static void main(String[] args) {
        /*
        Input Tree:
                5
            / \
            4   6
            \    \
            3    7
                    \
                    8
        */
        TreeNode root = new TreeNode(5);
        root.left = new TreeNode(4);
        root.left.right = new TreeNode(3);
        root.right = new TreeNode(6);
        root.right.right = new TreeNode(7);
        root.right.right.right = new TreeNode(8);

        TreeNode p = root.left;
        TreeNode q = root.left.right;

        TreeNode lca1 = lowestCommonAncestor(root, p, q);
        System.out.println("LCA of " + p.val + " and " + q.val + " is " + lca1.val);

        TreeNode x = root.right.right;
        TreeNode y = root.right.right.right;

        TreeNode lca2 = lowestCommonAncestor(root, x, y);
        System.out.println("LCA of " + x.val + " and " + y.val + " is " + lca2.val);
    }
}

Python

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

def lowest_common_ancestor(root, p, q):
    if not root or root == p or root == q:
        return root

    left_lca = lowest_common_ancestor(root.left, p, q)
    right_lca = lowest_common_ancestor(root.right, p, q)

    if left_lca and right_lca:
        return root

    return left_lca if left_lca else right_lca

# Driver Code
"""
Input Tree:
        5
    / \
    4   6
    \    \
    3    7
            \
            8
"""
root = TreeNode(5)
root.left = TreeNode(4)
root.left.right = TreeNode(3)
root.right = TreeNode(6)
root.right.right = TreeNode(7)
root.right.right.right = TreeNode(8)

p = root.left
q = root.left.right

lca1 = lowest_common_ancestor(root, p, q)
print(f"LCA of {p.val} and {q.val} is {lca1.val}")

x = root.right.right
y = root.right.right.right

lca2 = lowest_common_ancestor(root, x, y)
print(f"LCA of {x.val} and {y.val} is {lca2.val}")

using System;

public class TreeNode
{
    public int val;
    public TreeNode left, right;

    public TreeNode(int x)
    {
        val = x;
        left = right = null;
    }
}

public class BinaryTree
{
    public TreeNode LowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q)
    {
        if (root == null || root == p || root == q)
        {
            return root;
        }

        TreeNode leftLCA = LowestCommonAncestor(root.left, p, q);
        TreeNode rightLCA = LowestCommonAncestor(root.right, p, q);

        if (leftLCA != null && rightLCA != null)
        {
            return root;
        }

        return leftLCA != null ? leftLCA : rightLCA;
    }

    // Driver Code
    public static void Main()
    {
        /*
        Input Tree:
                5
            / \
            4   6
            \    \
            3    7
                    \
                    8
        */
        TreeNode root = new TreeNode(5);
        root.left = new TreeNode(4);
        root.left.right = new TreeNode(3);
        root.right = new TreeNode(6);
        root.right.right = new TreeNode(7);
        root.right.right.right = new TreeNode(8);

        TreeNode p = root.left;
        TreeNode q = root.left.right;

        BinaryTree binaryTree = new BinaryTree();
        TreeNode lca1 = binaryTree.LowestCommonAncestor(root, p, q);
        Console.WriteLine($"LCA of {p.val} and {q.val} is {lca1.val}");

        TreeNode x = root.right.right;
        TreeNode y = root.right.right.right;

        TreeNode lca2 = binaryTree.LowestCommonAncestor(root, x, y);
        Console.WriteLine($"LCA of {x.val} and {y.val} is {lca2.val}");
    }
}

Javascript

class TreeNode {
    constructor(x) {
        this.val = x;
        this.left = null;
        this.right = null;
    }
}

class BinaryTree {
    lowestCommonAncestor(root, p, q) {
        if (!root || root === p || root === q) {
            return root;
        }

        const leftLCA = this.lowestCommonAncestor(root.left, p, q);
        const rightLCA = this.lowestCommonAncestor(root.right, p, q);

        if (leftLCA && rightLCA) {
            return root;
        }

        return leftLCA || rightLCA;
    }
}

// Driver Code
/*
Input Tree:
        5
    / \
    4   6
    \    \
    3    7
            \
            8
*/
const root = new TreeNode(5);
root.left = new TreeNode(4);
root.left.right = new TreeNode(3);
root.right = new TreeNode(6);
root.right.right = new TreeNode(7);
root.right.right.right = new TreeNode(8);

const binaryTree = new BinaryTree();

const p = root.left;
const q = root.left.right;

const lca1 = binaryTree.lowestCommonAncestor(root, p, q);
console.log(`LCA of ${p.val} and ${q.val} is ${lca1.val}`);

const x = root.right.right;
const y = root.right.right.right;

const lca2 = binaryTree.lowestCommonAncestor(root, x, y);
console.log(`LCA of ${x.val} and ${y.val} is ${lca2.val}`);

Output

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

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

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

Tags:

#Accolite #Amazon #LCA #MAQ Software #Microsoft #Samsung #Synopsys #Binary Search Tree #DSA #Tree #Accolite #Amazon #MAQ Software #Microsoft #Samsung #Synopsys #Binary Search Tree #Tree

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

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