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# Print BST keys in the given range

Given two values k1 and k2 where k1 < k2 and a root pointer to a Binary Search Tree. The task is to print all the keys of the tree in the range k1 to k2 in increasing order.

Examples:

Input: k1 = 10 and k2 = 22

Output: 12, 20 and 22.
Explanation: The keys are 4, 8, 12, 20, and 22, So keys in range 10 to 22 is 12, 20 and 22.

Input: k1 = 1 and k2 = 10

Output: 8
Explanation: The key 8 is in the range 1 to 10

Approach: Below is the idea to solve the problem:

Traverse the tree in the inorder traversal. If the Binary search tree is traversed in inorder traversal the keys are traversed in increasing order. So while traversing the keys in the inorder traversal. If the key lies in the range print the key else skip the key.

Follow the below steps to Implement the idea:

• Run DFS on BST in in-order traversal starting from root keeping k1 and k2 as parameters.
• If the value of the root’s key is greater than k1, then recursively call in the left subtree i.e. if k1 < root->data call for root->left.
• If the value of the root’s key is in range, then print the root’s key i.e. if k1 <= root->data and k2 >= root->data print root->data.
• Recursively call for the right subtree i.e. root->right.

Below is the Implementation of the above approach:

## C++

 `// C++ program to print BST in given range` `#include` `using` `namespace` `std; `   `/* A tree node structure */` `class` `node ` `{ ` `    ``public``:` `    ``int` `data; ` `    ``node *left; ` `    ``node *right; ` `}; `   `/* The functions prints all the keys` `which in the given range [k1..k2]. ` `    ``The function assumes than k1 < k2 */` `void` `Print(node *root, ``int` `k1, ``int` `k2) ` `{ ` `    ``/* base case */` `    ``if` `( NULL == root ) ` `        ``return``; ` `    `  `    ``/* Since the desired o/p is sorted, ` `        ``recurse for left subtree first ` `        ``*/` `     `  `    ``Print(root->left, k1, k2); ` `    `  `    ``/* if root's data lies in range, ` `    ``then prints root's data */` `    ``if` `( k1 <= root->data && k2 >= root->data ) ` `        ``cout<data<<``" "``; ` `    `  `    ``/* recursively call the right subtree */` `   ``Print(root->right, k1, k2); ` `} `   `/* Utility function to create a new Binary Tree node */` `node* newNode(``int` `data) ` `{ ` `    ``node *temp = ``new` `node(); ` `    ``temp->data = data; ` `    ``temp->left = NULL; ` `    ``temp->right = NULL; ` `    `  `    ``return` `temp; ` `} `   `/* Driver code */` `int` `main() ` `{ ` `    ``node *root = ``new` `node(); ` `    ``int` `k1 = 10, k2 = 25; ` `    `  `    ``/* Constructing tree given` `    ``in the above figure */` `    ``root = newNode(20); ` `    ``root->left = newNode(8); ` `    ``root->right = newNode(22); ` `    ``root->left->left = newNode(4); ` `    ``root->left->right = newNode(12); ` `    `  `    ``Print(root, k1, k2); ` `    ``return` `0; ` `} `   `// This code is contributed by rathbhupendra`

## C

 `#include` `#include`   `/* A tree node structure */` `struct` `node` `{` `  ``int` `data;` `  ``struct` `node *left;` `  ``struct` `node *right;` `};`   `/* The functions prints all the keys which in ` `the given range [k1..k2]. The function assumes than k1 < k2 */` `void` `Print(``struct` `node *root, ``int` `k1, ``int` `k2)` `{` `   ``/* base case */` `   ``if` `( NULL == root )` `      ``return``;`   `   ``/* Since the desired o/p is sorted, recurse for left subtree first` `      ``If root->data is greater than k1, then only we can get o/p keys` `      ``in left subtree */` `   ``if` `( k1 < root->data )` `     ``Print(root->left, k1, k2);`   `   ``/* if root's data lies in range, then prints root's data */` `   ``if` `( k1 <= root->data && k2 >= root->data )` `     ``printf``(``"%d "``, root->data );`   `  ``/* recursively call the right subtree */` `   ``Print(root->right, k1, k2);` `}`   `/* Utility function to create a new Binary Tree node */` `struct` `node* newNode(``int` `data)` `{` `  ``struct` `node *temp = ``malloc``(``sizeof``(``struct` `node));` `  ``temp->data = data;` `  ``temp->left = NULL;` `  ``temp->right = NULL;`   `  ``return` `temp;` `}`   `/* Driver function to test above functions */` `int` `main()` `{` `  ``struct` `node *root = ``malloc``(``sizeof``(``struct` `node));` `  ``int` `k1 = 10, k2 = 25;`   `  ``/* Constructing tree given in the above figure */` `  ``root = newNode(20);` `  ``root->left = newNode(8);` `  ``root->right = newNode(22);` `  ``root->left->left = newNode(4);` `  ``root->left->right = newNode(12);`   `  ``Print(root, k1, k2);`   `  ``getchar``();` `  ``return` `0;` `}`     `// This code was modified by italovinicius18`

## Java

 `// Java program to print BST in given range`   `// A binary tree node` `class` `Node {`   `    ``int` `data;` `    ``Node left, right;`   `    ``Node(``int` `d) {` `        ``data = d;` `        ``left = right = ``null``;` `    ``}` `}`   `class` `BinaryTree {` `    `  `    ``static` `Node root;` `    `  `    ``/* The functions prints all the keys which in ` `     ``the given range [k1..k2]. The function assumes than k1 < k2 */` `    ``void` `Print(Node node, ``int` `k1, ``int` `k2) {` `        `  `        ``/* base case */` `        ``if` `(node == ``null``) {` `            ``return``;` `        ``}`   `        ``/* Since the desired o/p is sorted, recurse for left subtree first` `         ``If root->data is greater than k1, then only we can get o/p keys` `         ``in left subtree */` `        ``if` `(k1 < node.data) {` `            ``Print(node.left, k1, k2);` `        ``}`   `        ``/* if root's data lies in range, then prints root's data */` `        ``if` `(k1 <= node.data && k2 >= node.data) {` `            ``System.out.print(node.data + ``" "``);` `        ``}`   `        ``/* recursively call the right subtree  */` `         ``Print(node.right, k1, k2);` `    ``}` `    `  `    ``public` `static` `void` `main(String[] args) {` `        ``BinaryTree tree = ``new` `BinaryTree();` `        ``int` `k1 = ``10``, k2 = ``25``;` `        ``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.Print(root, k1, k2);` `    ``}` `}`   `// This code has been contributed by Mayank Jaiswal`

## Python3

 `# Python program to find BST keys in given range`   `# 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`   `# The function prints all the keys in the given range` `# [k1..k2]. Assumes that k1 < k2 ` `def` `Print``(root, k1, k2):` `    `  `    ``# Base Case` `    ``if` `root ``is` `None``:` `        ``return`   `    ``# Since the desired o/p is sorted, recurse for left` `    ``# subtree first. If root.data is greater than k1, then` `    ``# only we can get o/p keys in left subtree` `    ``if` `k1 < root.data :` `        ``Print``(root.left, k1, k2)`   `    ``# If root's data lies in range, then prints root's data` `    ``if` `k1 <``=` `root.data ``and` `k2 >``=` `root.data:` `        ``print` `(root.data,end``=``' '``)`   `    ``# recursively call the right subtree` `    ``Print``(root.right, k1, k2)`   `# Driver function to test above function` `k1 ``=` `10` `; k2 ``=` `25` `;` `root ``=` `Node(``20``)` `root.left ``=` `Node(``8``)` `root.right ``=` `Node(``22``)` `root.left.left ``=` `Node(``4``)` `root.left.right ``=` `Node(``12``)`   `Print``(root, k1, k2)`   `# This code is contributed by Nikhil Kumar Singh(nickzuck_007)`

## C#

 `using` `System;`   `// C# program to print BST in given range `   `// A binary tree node ` `public` `class` `Node` `{`   `    ``public` `int` `data;` `    ``public` `Node left, right;`   `    ``public` `Node(``int` `d)` `    ``{` `        ``data = d;` `        ``left = right = ``null``;` `    ``}` `}`   `public` `class` `BinaryTree` `{`   `    ``public` `static` `Node root;`   `    ``/* The functions prints all the keys which in the ` `     ``given range [k1..k2]. The function assumes than k1 < k2 */` `    ``public` `virtual` `void` `Print(Node node, ``int` `k1, ``int` `k2)` `    ``{`   `        ``/* base case */` `        ``if` `(node == ``null``)` `        ``{` `            ``return``;` `        ``}`   `        ``/* Since the desired o/p is sorted, recurse for left subtree first ` `         ``If root->data is greater than k1, then only we can get o/p keys ` `         ``in left subtree */` `        ``if` `(k1 < node.data)` `        ``{` `            ``Print(node.left, k1, k2);` `        ``}`   `        ``/* if root's data lies in range, then prints root's data */` `        ``if` `(k1 <= node.data && k2 >= node.data)` `        ``{` `            ``Console.Write(node.data + ``" "``);` `        ``}`   `        ``/* recursively call the right subtree */` `            ``Print(node.right, k1, k2);` `    ``}`   `    ``public` `static` `void` `Main(``string``[] args)` `    ``{` `        ``BinaryTree tree = ``new` `BinaryTree();` `        ``int` `k1 = 10, k2 = 25;` `        ``BinaryTree.root = ``new` `Node(20);` `        ``BinaryTree.root.left = ``new` `Node(8);` `        ``BinaryTree.root.right = ``new` `Node(22);` `        ``BinaryTree.root.left.left = ``new` `Node(4);` `        ``BinaryTree.root.left.right = ``new` `Node(12);`   `        ``tree.Print(root, k1, k2);` `    ``}` `}`   `// This code is contributed by Shrikant13`

## Javascript

 ``

Output

`12 20 22 `

Time Complexity: O(N), where N is the total number of keys in the tree, A single traversal of the tree is needed.
Auxiliary Space: O(H). where H is the height of the tree in recursion call stack

Approach 2: Using Morris Inorder Traversal

In this approach, we use Morris Inorder Traversal to traverse the tree without using a stack or recursion. In Morris Inorder Traversal, we make use of threaded binary trees to find the predecessor of the current node. We start with the root node and traverse to the leftmost node of the subtree rooted at the current node. For each node in the subtree, we check if it lies within the given range and print its key. If the node has a right child, we find its inorder successor using the threaded binary tree and update the pointers to create a threaded binary tree.

• If the given root node is NULL, return.
• Initialize a current pointer to point to the root node.
• While the current pointer is not NULL, do the following:
•  If the left child of the current node is NULL:
• If the key of the current node is within the given range, print the key.
•  Move to the right child of the current node.
•  If the left child of the current node is not NULL:
•  Find the rightmost node in the left subtree of the current node.
•  If the right child of the rightmost node is NULL, set it to point to the current node and move to the left child of the current node.
•  If the right child of the rightmost node is not NULL, set it to NULL, check if the key of the current node is within the given range, and move to the right child of the current node.

Here’s the C++ code for this approach:

## C++

 `#include ` `using` `namespace` `std; `   `struct` `Node ` `{ ` `    ``int` `key; ` `    ``Node *left, *right; ` `}; `   `Node* newNode(``int` `item) ` `{ ` `    ``Node* temp = ``new` `Node; ` `    ``temp->key = item; ` `    ``temp->left = temp->right = NULL; ` `    ``return` `temp; ` `} `   `void` `printRange(Node* root, ``int` `lower, ``int` `upper) ` `{ ` `    ``if` `(root == NULL) ` `        ``return``; ` `    `  `    ``Node* curr = root; ` `    ``while` `(curr != NULL) ` `    ``{ ` `        ``if` `(curr->left == NULL) ` `        ``{ ` `            ``if` `(curr->key >= lower && curr->key <= upper) ` `                ``cout << curr->key << ``" "``; ` `            ``curr = curr->right; ` `        ``} ` `        ``else` `        ``{ ` `            ``Node* pre = curr->left; ` `            ``while` `(pre->right != NULL && pre->right != curr) ` `                ``pre = pre->right; `   `            ``if` `(pre->right == NULL) ` `            ``{ ` `                ``pre->right = curr; ` `                ``curr = curr->left; ` `            ``} ` `            ``else` `            ``{ ` `                ``pre->right = NULL; ` `                ``if` `(curr->key >= lower && curr->key <= upper) ` `                    ``cout << curr->key << ``" "``; ` `                ``curr = curr->right; ` `            ``} ` `        ``} ` `    ``} ` `} `   `int` `main() ` `{ ` `    ``Node* root = newNode(20); ` `    ``root->left = newNode(8); ` `    ``root->right = newNode(22); ` `    ``root->left->left = newNode(4); ` `    ``root->right->left = newNode(12); `   `    ``int` `lower = 10, upper = 25; ` `    ``printRange(root, lower, upper); ` `    ``return` `0; ` `} `

## Python3

 `class` `Node:` `    ``def` `__init__(``self``, item):` `        ``self``.key ``=` `item` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None`   `def` `newNode(item):` `    ``temp ``=` `Node(item)` `    ``return` `temp`   `def` `printRange(root, lower, upper):` `    ``if` `root ``is` `None``:` `        ``return`   `    ``curr ``=` `root` `    ``while` `curr ``is` `not` `None``:` `        ``if` `curr.left ``is` `None``:` `            ``if` `curr.key >``=` `lower ``and` `curr.key <``=` `upper:` `                ``print``(curr.key, end``=``' '``)` `            ``curr ``=` `curr.right` `        ``else``:` `            ``pre ``=` `curr.left` `            ``while` `pre.right ``is` `not` `None` `and` `pre.right !``=` `curr:` `                ``pre ``=` `pre.right`   `            ``if` `pre.right ``is` `None``:` `                ``pre.right ``=` `curr` `                ``curr ``=` `curr.left` `            ``else``:` `                ``pre.right ``=` `None` `                ``if` `curr.key >``=` `lower ``and` `curr.key <``=` `upper:` `                    ``print``(curr.key, end``=``' '``)` `                ``curr ``=` `curr.right`   `root ``=` `newNode(``20``)` `root.left ``=` `newNode(``8``)` `root.right ``=` `newNode(``22``)` `root.left.left ``=` `newNode(``4``)` `root.right.left ``=` `newNode(``12``)`   `lower ``=` `10` `upper ``=` `25` `printRange(root, lower, upper)`

Output

`20 12 22 `

Time Complexity: O(N), where N is the total number of keys in the tree, A single traversal of the tree is needed.
Auxiliary Space: O(1).