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. In the diagram above, the lowest common ancestor of the nodes 4 and 6 is the node 3. Node 3 is the lowest node which has nodes and as descendants. 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. 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 and .
Solution :
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;
}
In Java :
/* Node is defined as :
class Node
int data;
Node left;
Node right;
*/
static Node lca(Node root,int v1,int v2)
{
Queue<String> v1path = new LinkedList<String>();
Queue<String> v2path = new LinkedList<String>();
String left = "l";
String right = "r";
Node rootcopy1 = new Node();
rootcopy1 = root;
Node rootcopy2 = new Node();
rootcopy2 = root;
while(rootcopy1 != null) //finding v1's path
{
if(v1 == rootcopy1.data) //found it, no need to continue
{
break;
}
else if(v1 < rootcopy1.data)
{
v1path.add(left);
rootcopy1 = rootcopy1.left;
}
else
{
v1path.add(right);
rootcopy1 = rootcopy1.right;
}
}
while(rootcopy2 != null) //finding v2's path
{
if(v2 == rootcopy2.data) //found it, no need to continue
{
break;
}
else if(v2 < rootcopy2.data)
{
v2path.add(left);
rootcopy2 = rootcopy2.left;
}
else
{
v2path.add(right);
rootcopy2 = rootcopy2.right;
}
}
rootcopy1 = root; //need to travel along tree down from root again to find LCA to return
while(v1path.peek() != null && v2path.peek() != null) //comparing paths taken to find v1 and v2 to determine LCA
{
if(v1path.peek() == v2path.peek()) //if paths take same branch, LCA is farther down from here, keep going
{
if(v1path.peek() == "l") //take left
{
v1path.remove();
v2path.remove();
rootcopy1 = rootcopy1.left;
}
else //take right
{
v1path.remove();
v2path.remove();
rootcopy1 = rootcopy1.right;
}
}
else //this is the LCA since the paths split here
{
break;
}
}
return rootcopy1;
}
In python3 :
# Enter your code here. Read input from STDIN. Print output to STDOUT
'''
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
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;
}
View More Similar Problems
Tree: Postorder Traversal
Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the
View Solution →Tree: Inorder Traversal
In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your $inOrder* func
View Solution →Tree: Height of a Binary Tree
The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary
View Solution →Tree : Top View
Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.
View Solution →Tree: Level Order Traversal
Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F
View Solution →Binary Search Tree : Insertion
You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <
View Solution →