Binary Search Tree : Lowest Common Ancestor
Problem Statement :
You are given pointer to the root of the binary search tree and two values v1 and v2. You need to return the lowest common ancestor (LCA) of v1 and v2 in the binary search tree. Function Description Complete the function lca in the editor below. It should return a pointer to the lowest common ancestor node of the two values given. lca has the following parameters: - root: a pointer to the root node of a binary search tree - v1: a node.data value - v2: a node.data value Input Format The first line contains an integer, n, the number of nodes in the tree. The second line contains n space-separated integers representing node.data values. The third line contains two space-separated integers, v1 and v2. To use the test data, you will have to create the binary search tree yourself. Here on the platform, the tree will be created for you. Constraints 1 <= n, node.data <= 25 1 <= v1, v2 <= 25 v1 =/ v2 The tree will contain nodes with data equal to v1 and v2. Output Format Return the a pointer to the node that is the lowest common ancestor of v1 and v2.
Solution :
Solution in C :
In C :
/* you only have to complete the function given below.
node is defined as
struct node {
int data;
struct node *left;
struct node *right;
};
*/
struct node *lca( struct node *root, int v1, int v2 ) {
if( root->data == v1 || root->data == v2 )
return root;
if( root->data > v1 && root->data > v2 )
return lca(root->left, v1, v2);
if( root->data < v1 && root->data < v2 )
return lca(root->right, v1, v2);
return root;
}
Solution in C++ :
In C++ :
/*
Node is defined as
typedef struct node
{
int data;
node * left;
node * right;
}node;
*/
bool isPresent(struct node* tree, int k)
{
if(tree==NULL)
return false;
if(tree->data==k)
return true;
return isPresent(tree->left,k) || isPresent(tree->right,k);
}
node * lca(node * tree, int m,int n)
{
if(tree==NULL)
return NULL;
if(tree->data>m && tree->data>n)
return lca(tree->left,m,n);
if(tree->data<m && tree->data<n)
return lca(tree->right,m,n);
if(tree->data<m && tree->data>n)
{
if(isPresent(tree->left,n) && isPresent(tree->right,m))
{
return tree;
}
else
return NULL;
}
if(tree->data>m && tree->data<n)
{
if(isPresent(tree->left,m) && isPresent(tree->right,n))
{
return tree;
}
else
return NULL;
}
if(tree->data==m)
{
if(tree->data<n)
{
if(isPresent(tree->right,n))
return tree;
else
return NULL;
}
if(tree->data>n)
{
if(isPresent(tree->left,n))
return tree;
else
return NULL;
}
if(tree->data==n)
return tree;
}
if(tree->data==n)
{
if(tree->data<m)
{
if(isPresent(tree->right,m))
return tree;
else
return NULL;
}
if(tree->data>m)
{
if(isPresent(tree->left,m))
return tree;
else
return NULL;
}
if(tree->data==m)
return tree;
}
return NULL;
}
Solution in Java :
In Java :
/* Node is defined as :
class Node
int data;
Node left;
Node right;
*/
static Node lca(Node root,int v1,int v2)
{
int min=Math.min(v1,v2);
int max=Math.max(v1,v2);
if(root.data>=min && root.data<=max){
return root;
}else if(root.data>max && root.data>max){
return lca(root.left,v1,v2);
}else{
return lca(root.right,v1,v2);
}
}
Solution in Python :
In Python3 :
'''
class Node:
def __init__(self,info):
self.info = info
self.left = None
self.right = None
// this is a node of the tree , which contains info as data, left , right
'''
def lca(root, v1, v2):
node=root
if node is None:
return None
if node.info>v1 and node.info>v2:
return lca(node.left,v1,v2)
elif node.info<v1 and node.info < v2:
return lca(node.right,v1,v2)
else:
return node
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