Find median of BST in O(n) time and O(1) space

Given a Binary Search Tree, find median of it.

If no. of nodes are even: then median = ((n/2th node + (n+1)/2th node) /2
If no. of nodes are odd : then median = (n+1)/2th node.

For example, median of below BST is 12.



More Examples:

 Given BST(with odd no. of nodes) is : 
                    6
                 /    \
                3       8
              /   \    /  \
             1     4  7    9

Inorder of Given BST will be : 1, 3, 4, 6, 7, 8, 9
So, here median will 6.

Given BST(with even no. of nodes) is :  
                    6
                 /    \
                3       8
              /   \    /  
             1     4  7    

Inorder of Given BST will be : 1, 3, 4, 6, 7, 8
So, here median will  (4+6)/2 = 5.

Asked in : Google

To find the median, we need to find the Inorder of the BST because its Inorder will be in sorted order and then find the median i.e.

The idea is based on K’th smallest element in BST using O(1) Extra Space

The task is very simple if we are allowed to use extra space but Inorder traversal using recursion and stack both uses Space which is not allowed here. So, the solution is to do Morris Inorder traversal as it doesn’t require any extra space.

Implementation:
1- Count the no. of nodes in the given BST
   using Morris Inorder Traversal.
2- Then Perform Morris Inorder traversal one 
   more time by counting nodes and by checking if 
   count is equal to the median point.
   To consider even no. of nodes an extra pointer
   pointing to the previous node is used.

C++

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/* C++ program to find the median of BST in O(n)
   time and O(1) space*/
#include<iostream>
using namespace std;
  
/* A binary search tree Node has data, pointer
   to left child and a pointer to right child */
struct Node
{
    int data;
    struct Node* left, *right;
};
  
// A utility function to create a new BST node
struct Node *newNode(int item)
{
    struct Node *temp =  new Node;
    temp->data = item;
    temp->left = temp->right = NULL;
    return temp;
}
  
/* A utility function to insert a new node with
   given key in BST */
struct Node* insert(struct Node* node, int key)
{
    /* If the tree is empty, return a new node */
    if (node == NULL) return newNode(key);
  
    /* Otherwise, recur down the tree */
    if (key < node->data)
        node->left  = insert(node->left, key);
    else if (key > node->data)
        node->right = insert(node->right, key);
  
    /* return the (unchanged) node pointer */
    return node;
}
  
/* Function to count nodes in a  binary search tree
   using Morris Inorder traversal*/
int counNodes(struct Node *root)
{
    struct Node *current, *pre;
  
    // Initialise count of nodes as 0
    int count = 0;
  
    if (root == NULL)
        return count;
  
    current = root;
    while (current != NULL)
    {
        if (current->left == NULL)
        {
            // Count node if its left is NULL
            count++;
  
            // Move to its right
            current = current->right;
        }
        else
        {
            /* Find the inorder predecessor of current */
            pre = current->left;
  
            while (pre->right != NULL &&
                   pre->right != current)
                pre = pre->right;
  
            /* Make current as right child of its
               inorder predecessor */
            if(pre->right == NULL)
            {
                pre->right = current;
                current = current->left;
            }
  
            /* Revert the changes made in if part to
               restore the original tree i.e., fix
               the right child of predecssor */
            else
            {
                pre->right = NULL;
  
                // Increment count if the current
                // node is to be visited
                count++;
                current = current->right;
            } /* End of if condition pre->right == NULL */
        } /* End of if condition current->left == NULL*/
    } /* End of while */
  
    return count;
}
  
  
/* Function to find median in O(n) time and O(1) space
   using Morris Inorder traversal*/
int findMedian(struct Node *root)
{
   if (root == NULL)
        return 0;
  
    int count = counNodes(root);
    int currCount = 0;
    struct Node *current = root, *pre, *prev;
  
    while (current != NULL)
    {
        if (current->left == NULL)
        {
            // count current node
            currCount++;
  
            // check if current node is the median
            // Odd case
            if (count % 2 != 0 && currCount == (count+1)/2)
                return prev->data;
  
            // Even case
            else if (count % 2 == 0 && currCount == (count/2)+1)
                return (prev->data + current->data)/2;
  
            // Update prev for even no. of nodes
            prev = current;
  
            //Move to the right
            current = current->right;
        }
        else
        {
            /* Find the inorder predecessor of current */
            pre = current->left;
            while (pre->right != NULL && pre->right != current)
                pre = pre->right;
  
            /* Make current as right child of its inorder predecessor */
            if (pre->right == NULL)
            {
                pre->right = current;
                current = current->left;
            }
  
            /* Revert the changes made in if part to restore the original
              tree i.e., fix the right child of predecssor */
            else
            {
                pre->right = NULL;
  
                prev = pre;
  
                // Count current node
                currCount++;
  
                // Check if the current node is the median
                if (count % 2 != 0 && currCount == (count+1)/2 )
                    return current->data;
  
                else if (count%2==0 && currCount == (count/2)+1)
                    return (prev->data+current->data)/2;
  
                // update prev node for the case of even
                // no. of nodes
                prev = current;
                current = current->right;
  
            } /* End of if condition pre->right == NULL */
        } /* End of if condition current->left == NULL*/
    } /* End of while */
}
  
/* Driver program to test above functions*/
int main()
{
  
    /* Let us create following BST
                  50
               /     \
              30      70
             /  \    /  \
           20   40  60   80 */
    struct Node *root = NULL;
    root = insert(root, 50);
    insert(root, 30);
    insert(root, 20);
    insert(root, 40);
    insert(root, 70);
    insert(root, 60);
    insert(root, 80);
  
    cout << "\nMedian of BST is "
         << findMedian(root);
    return 0;
}

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Java

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/* Java program to find the median of BST in O(n) 
time and O(1) space*/
class GfG { 
  
/* A binary search tree Node has data, pointer 
to left child and a pointer to right child */
static class Node 
    int data; 
    Node left, right; 
}
  
// A utility function to create a new BST node 
static Node newNode(int item) 
    Node temp = new Node(); 
    temp.data = item; 
    temp.left = null;
    temp.right = null
    return temp; 
  
/* A utility function to insert a new node with 
given key in BST */
static Node insert(Node node, int key) 
    /* If the tree is empty, return a new node */
    if (node == null) return newNode(key); 
  
    /* Otherwise, recur down the tree */
    if (key < node.data) 
        node.left = insert(node.left, key); 
    else if (key > node.data) 
        node.right = insert(node.right, key); 
  
    /* return the (unchanged) node pointer */
    return node; 
  
/* Function to count nodes in a binary search tree 
using Morris Inorder traversal*/
static int counNodes(Node root) 
    Node current, pre; 
  
    // Initialise count of nodes as 0 
    int count = 0
  
    if (root == null
        return count; 
  
    current = root; 
    while (current != null
    
        if (current.left == null
        
            // Count node if its left is NULL 
            count++; 
  
            // Move to its right 
            current = current.right; 
        
        else
        
            /* Find the inorder predecessor of current */
            pre = current.left; 
  
            while (pre.right != null && 
                pre.right != current) 
                pre = pre.right; 
  
            /* Make current as right child of its 
            inorder predecessor */
            if(pre.right == null
            
                pre.right = current; 
                current = current.left; 
            
  
            /* Revert the changes made in if part to 
            restore the original tree i.e., fix 
            the right child of predecssor */
            else
            
                pre.right = null
  
                // Increment count if the current 
                // node is to be visited 
                count++; 
                current = current.right; 
            } /* End of if condition pre->right == NULL */
        } /* End of if condition current->left == NULL*/
    } /* End of while */
  
    return count; 
  
  
/* Function to find median in O(n) time and O(1) space 
using Morris Inorder traversal*/
static int findMedian(Node root) 
if (root == null
        return 0
  
    int count = counNodes(root); 
    int currCount = 0
    Node current = root, pre = null, prev = null
  
    while (current != null
    
        if (current.left == null
        
            // count current node 
            currCount++; 
  
            // check if current node is the median 
            // Odd case 
            if (count % 2 != 0 && currCount == (count+1)/2
                return prev.data; 
  
            // Even case 
            else if (count % 2 == 0 && currCount == (count/2)+1
                return (prev.data + current.data)/2
  
            // Update prev for even no. of nodes 
            prev = current; 
  
            //Move to the right 
            current = current.right; 
        
        else
        
            /* Find the inorder predecessor of current */
            pre = current.left; 
            while (pre.right != null && pre.right != current) 
                pre = pre.right; 
  
            /* Make current as right child of its inorder predecessor */
            if (pre.right == null
            
                pre.right = current; 
                current = current.left; 
            
  
            /* Revert the changes made in if part to restore the original 
            tree i.e., fix the right child of predecssor */
            else
            
                pre.right = null
  
                prev = pre; 
  
                // Count current node 
                currCount++; 
  
                // Check if the current node is the median 
                if (count % 2 != 0 && currCount == (count+1)/2
                    return current.data; 
  
                else if (count%2==0 && currCount == (count/2)+1
                    return (prev.data+current.data)/2
  
                // update prev node for the case of even 
                // no. of nodes 
                prev = current; 
                current = current.right; 
  
            } /* End of if condition pre->right == NULL */
        } /* End of if condition current->left == NULL*/
    } /* End of while */
    return -1;
  
/* Driver code*/
public static void main(String[] args) 
  
    /* Let us create following BST 
                50 
            / \ 
            30 70 
            / \ / \ 
        20 40 60 80 */
    Node root = null
    root = insert(root, 50); 
    insert(root, 30); 
    insert(root, 20); 
    insert(root, 40); 
    insert(root, 70); 
    insert(root, 60); 
    insert(root, 80); 
  
    System.out.println("Median of BST is " + findMedian(root)); 
}
  
// This code is contributed by prerna saini.

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Python3

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# Python program to find closest 
# value in Binary search Tree
  
_MIN=-2147483648
_MAX=2147483648
  
# Helper function that allocates  
# a new node with the given data  
# and None left and right poers.                                     
class newNode: 
  
    # Constructor to create a new node 
    def __init__(self, data): 
        self.data = data 
        self.left = None
        self.right = None
  
""" A utility function to insert a new node with
given key in BST """
def insert(node,key):
  
    """ If the tree is empty, return a new node """
    if (node == None):
        return newNode(key)
  
    """ Otherwise, recur down the tree """
    if (key < node.data):
        node.left = insert(node.left, key)
    elif (key > node.data):
        node.right = insert(node.right, key)
  
    """ return the (unchanged) node pointer """
    return node
  
  
""" Function to count nodes in 
    a binary search tree using
     Morris Inorder traversal"""
def counNodes(root):
  
    # Initialise count of nodes as 0
    count = 0
  
    if (root == None):
        return count
  
    current = root
    while (current != None):
      
        if (current.left == None):
          
            # Count node if its left is None
            count+=1
  
            # Move to its right
            current = current.right
          
        else:     
            """ Find the inorder predecessor of current """
            pre = current.left
  
            while (pre.right != None and 
                    pre.right != current):
                pre = pre.right
  
            """ Make current as right child of its
            inorder predecessor """
            if(pre.right == None):
              
                pre.right = current
                current = current.left
            else:
              
                pre.right = None
  
                # Increment count if the current
                # node is to be visited
                count += 1
                current = current.right
  
    return count
  
  
  
""" Function to find median in
    O(n) time and O(1) space
    using Morris Inorder traversal"""
def findMedian(root):
    if (root == None):
        return 0
    count = counNodes(root)
    currCount = 0
    current = root
  
    while (current != None):
      
        if (current.left == None):
          
            # count current node
            currCount += 1
  
            # check if current node is the median
            # Odd case
            if (count % 2 != 0 and 
                currCount == (count + 1)//2):
                return prev.data
  
            # Even case
            elif (count % 2 == 0 and 
                    currCount == (count//2)+1):
                return (prev.data + current.data)//2
  
            # Update prev for even no. of nodes
            prev = current
  
            #Move to the right
            current = current.right
          
        else:
          
            """ Find the inorder predecessor of current """
            pre = current.left
            while (pre.right != None and 
                    pre.right != current):
                pre = pre.right
  
            """ Make current as right child
                of its inorder predecessor """
            if (pre.right == None):
              
                pre.right = current
                current = current.left
            else:
              
                pre.right = None
  
                prev = pre
  
                # Count current node
                currCount+= 1
  
                # Check if the current node is the median
                if (count % 2 != 0 and 
                    currCount == (count + 1) // 2 ):
                    return current.data
  
                elif (count%2 == 0 and 
                    currCount == (count // 2) + 1):
                    return (prev.data+current.data)//2
  
                # update prev node for the case of even
                # no. of nodes
                prev = current
                current = current.right
  
          
# Driver Code 
if __name__ == '__main__':
  
    """ Constructed binary tree is
        50
        / \
    30 70
    / \ / \
    20 40 60 80 """
      
    root = newNode(50
    insert(root, 30)
    insert(root, 20)
    insert(root, 40)
    insert(root, 70)
    insert(root, 60)
    insert(root, 80)
    print("Median of BST is ",findMedian(root))
  
  
  
# This code is contributed
# Shubham Singh(SHUBHAMSINGH10)

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C#

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/* C# program to find the median of BST in O(n) 
time and O(1) space*/
using System;
class GfG 
  
/* A binary search tree Node has data, pointer 
to left child and a pointer to right child */
public class Node 
    public int data; 
    public Node left, right; 
  
// A utility function to create a new BST node 
static Node newNode(int item) 
    Node temp = new Node(); 
    temp.data = item; 
    temp.left = null
    temp.right = null
    return temp; 
  
/* A utility function to insert a new node with 
given key in BST */
static Node insert(Node node, int key) 
    /* If the tree is empty, return a new node */
    if (node == null) return newNode(key); 
  
    /* Otherwise, recur down the tree */
    if (key < node.data) 
        node.left = insert(node.left, key); 
    else if (key > node.data) 
        node.right = insert(node.right, key); 
  
    /* return the (unchanged) node pointer */
    return node; 
  
/* Function to count nodes in a binary search tree 
using Morris Inorder traversal*/
static int counNodes(Node root) 
    Node current, pre; 
  
    // Initialise count of nodes as 0 
    int count = 0; 
  
    if (root == null
        return count; 
  
    current = root; 
    while (current != null
    
        if (current.left == null
        
            // Count node if its left is NULL 
            count++; 
  
            // Move to its right 
            current = current.right; 
        
        else
        
            /* Find the inorder predecessor of current */
            pre = current.left; 
  
            while (pre.right != null && 
                pre.right != current) 
                pre = pre.right; 
  
            /* Make current as right child of its 
            inorder predecessor */
            if(pre.right == null
            
                pre.right = current; 
                current = current.left; 
            
  
            /* Revert the changes made in if part to 
            restore the original tree i.e., fix 
            the right child of predecssor */
            else
            
                pre.right = null
  
                // Increment count if the current 
                // node is to be visited 
                count++; 
                current = current.right; 
            } /* End of if condition pre->right == NULL */
        } /* End of if condition current->left == NULL*/
    } /* End of while */
  
    return count; 
  
  
/* Function to find median in O(n) time and O(1) space 
using Morris Inorder traversal*/
static int findMedian(Node root) 
if (root == null
        return 0; 
  
    int count = counNodes(root); 
    int currCount = 0; 
    Node current = root, pre = null, prev = null
  
    while (current != null
    
        if (current.left == null
        
            // count current node 
            currCount++; 
  
            // check if current node is the median 
            // Odd case 
            if (count % 2 != 0 && currCount == (count+1)/2) 
                return prev.data; 
  
            // Even case 
            else if (count % 2 == 0 && currCount == (count/2)+1) 
                return (prev.data + current.data)/2; 
  
            // Update prev for even no. of nodes 
            prev = current; 
  
            //Move to the right 
            current = current.right; 
        
        else
        
            /* Find the inorder predecessor of current */
            pre = current.left; 
            while (pre.right != null && pre.right != current) 
                pre = pre.right; 
  
            /* Make current as right child of its inorder predecessor */
            if (pre.right == null
            
                pre.right = current; 
                current = current.left; 
            
  
            /* Revert the changes made in if part to restore the original 
            tree i.e., fix the right child of predecssor */
            else
            
                pre.right = null
  
                prev = pre; 
  
                // Count current node 
                currCount++; 
  
                // Check if the current node is the median 
                if (count % 2 != 0 && currCount == (count+1)/2 ) 
                    return current.data; 
  
                else if (count % 2 == 0 && currCount == (count/2)+1) 
                    return (prev.data + current.data)/2; 
  
                // update prev node for the case of even 
                // no. of nodes 
                prev = current; 
                current = current.right; 
  
            } /* End of if condition pre->right == NULL */
        } /* End of if condition current->left == NULL*/
    } /* End of while */
    return -1; 
  
/* Driver code*/
public static void Main(String []args) 
  
    /* Let us create following BST 
                50 
            / \ 
            30 70 
            / \ / \ 
        20 40 60 80 */
    Node root = null
    root = insert(root, 50); 
    insert(root, 30); 
    insert(root, 20); 
    insert(root, 40); 
    insert(root, 70); 
    insert(root, 60); 
    insert(root, 80); 
  
    Console.WriteLine("Median of BST is " + findMedian(root)); 
  
// This code is contributed by Arnab Kundu

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Output:

Median of BST is 50

Reference:
https://www.careercup.com/question?id=4882624968392704

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