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# Find median of BST

• Difficulty Level : Hard
• Last Updated : 01 Sep, 2022

Given a Binary Search Tree, find the median of it.

• If number of nodes are even: then median = ((n/2th node + ((n)/2th+1) node) /2
• If number of nodes are odd: then median = (n+1)/2th node.

Examples:

Input: Example of BST

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

Given BST: Output: median of Below BST is 12

## Median Of BST using Morris Inorder Traversal:

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 to traversal using recursion and stack both use Space which is not allowed here. So, the solution is to do Morris Inorder traversal as it doesn’t require extra space.

Follow the steps mentioned below to implement the idea:

• Count the number of nodes in the given BST using Morris Inorder Traversal.
• Then perform Morris Inorder traversal one more time by counting nodes and by checking if the count is equal to the median point.
• To consider even no. of nodes, an extra pointer pointing to the previous node is used.

Below is the implementation of the above approach:

## C++

 `/* C++ program to find the median of BST in O(n)` `   ``time and O(1) space*/` `#include` `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 predecessor */` `            ``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` `current->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 predecessor */` `            ``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;` `}`

## Java

 `/* 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 predecessor */` `            ``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 predecessor */` `            ``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.`

## Python3

 `# 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 pointers.                                     ` `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)`

## C#

 `/* 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 predecessor */` `            ``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 predecessor */` `            ``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`

## Javascript

 ``

Output

`Median of BST is 50`

Time Complexity: O(N)
Auxiliary Space: O(1)

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