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Given a binary tree, write a function to get the maximum width of the given tree. Width of a tree is maximum of widths of all levels.Let us consider the below example tree. &nbsp;Input: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 1 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;/ &nbsp; \ &nbsp; &nbsp; &nbsp; 2 &nbsp; &nbsp; &nbsp;3 &nbsp; &nbsp;/ &nbsp; \ &nbsp; &nbsp; &nbsp; \ 4 &nbsp; &nbsp; 5 &nbsp; &nbsp; &nbsp; 8&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;/ &nbsp; &nbsp; \ &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 6 &nbsp; &nbsp; &nbsp; &nbsp;7 Output: &nbsp;3 Explanation: For the above tree,&nbsp; width of level 1 is 1,&nbsp; width of level 2 is 2,&nbsp; width of level 3 is 3&nbsp; width of level 4 is 2.&nbsp; So the maximum width of the tree is 3.Maximum width of a binary tree: <a href="https://www.geeksforgeeks.org/maximum-width-of-a-binary-tree/">https://www.geeksforgeeks.org/maximum-width-of-a-binary-tree/</a>

Maximum width of a binary tree

Given a root of a tree, and an integer k. Print all the nodes which are at k distance from root.&nbsp; For example, in the below tree, 4, 5 &amp; 8 are at distance 2 from root.&nbsp; &nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;1 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;/ &nbsp; \ &nbsp; &nbsp; &nbsp; &nbsp;2 &nbsp; &nbsp; &nbsp;3 &nbsp; &nbsp; &nbsp;/ &nbsp;\ &nbsp; &nbsp;/ &nbsp; &nbsp;4 &nbsp; &nbsp; 5 &nbsp;8Print nodes at k distance from root: <a href="https://www.geeksforgeeks.org/print-nodes-at-k-distance-from-root/">https://www.geeksforgeeks.org/print-nodes-at-k-distance-from-root/</a>

Print nodes at k distance from root

In Binary Tree, Inorder successor of a node is the next node in Inorder traversal of the Binary Tree. Inorder Successor is NULL for the last node in Inorder traversal.In Binary Search Tree, Inorder Successor of an input node can also be defined as the node with the smallest key greater than the key of the input node. So, it is sometimes important to find next node in sorted order.Inorder Successor in Binary Search Tree: <a href="https://www.geeksforgeeks.org/inorder-successor-in-binary-search-tree/">https://www.geeksforgeeks.org/inorder-successor-in-binary-search-tree/</a>

Inorder Successor in Binary Search Tree

Given values of two values n1 and n2 in a Binary Search Tree, find the Lowest Common Ancestor (LCA). You may assume that both the values exist in the tree.&nbsp; Input: LCA of 10 and 14 Output: &nbsp;12 Explanation: 12 is the closest node to both 10 and 14&nbsp; Algorithm: &nbsp;Create a recursive function that takes a node and the two values n1 and n2. If the value of the current node is less than both n1 and n2, then LCA lies in the right subtree. Call the recursive function for the right subtree. If the value of the current node is greater than both n1 and n2, then LCA lies in the left subtree. Call the recursive function for the left subtree. If both the above cases are false then return the current node as LCA.&nbsp;Lowest Common Ancestor in a Binary Search Tree &nbsp;: <a href="https://www.geeksforgeeks.org/lowest-common-ancestor-in-a-binary-search-tree/">https://www.geeksforgeeks.org/lowest-common-ancestor-in-a-binary-search-tree/</a>

Lowest Common Ancestor in a Binary Search Tree

Given a binary tree and a number, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals the given number. Return false if no such path can be found.&nbsp;Example:Input: &nbsp; &nbsp; &nbsp;10 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;Sum = 23 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;/ &nbsp; &nbsp; \ &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 8 &nbsp; &nbsp; &nbsp; &nbsp;2 &nbsp; &nbsp; &nbsp; &nbsp;/ &nbsp; &nbsp;\ &nbsp; &nbsp; / &nbsp; &nbsp; &nbsp;3 &nbsp; &nbsp; &nbsp;5 &nbsp;2Output: True&nbsp; Explanation: &nbsp;Root to leaf path sum, existing in this tree are 21 –&gt; 10 – 8 – 3&nbsp; 23 –&gt; 10 – 8 – 5&nbsp; 14 –&gt; 10 – 2 – 2 So it is possible to get sum = 23Follow the given steps to solve the problem using the above approach:<ul><li>Recursively move to the left and right subtree and at each call decrease the sum by the value of the current node</li><li> If at any level the current node is a leaf node and the remaining sum is equal to zero then return true</li></ul>Related Articles : <a href="https://www.geeksforgeeks.org/root-to-leaf-path-sum-equal-to-a-given-number/">https://www.geeksforgeeks.org/root-to-leaf-path-sum-equal-to-a-given-number/</a>

Root to leaf path sum equal to a given number

Given two values k1 and k2 (where k1 &lt; k2) and a root pointer to a Binary Search Tree. Print all the keys of the tree in range k1 to k2. i.e. print all x such that k1&lt;=x&lt;=k2 and x is a key of given BST. Print all the keys in increasing order.&nbsp; Example:Graph: 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.Print BST keys in the given range: <a href="https://www.geeksforgeeks.org/print-bst-keys-in-the-given-range/">https://www.geeksforgeeks.org/print-bst-keys-in-the-given-range/</a>

Print BST keys in the given range

AVL Tree is a self-balancing Binary Search Tree where the height difference between left and right subtrees is at most one. Named after its inventors, it ensures O(Log n) time complexity for search, insertion, and deletion. AVL Trees perform rotations to maintain balance and allow efficient operations. They are commonly used in scenarios requiring predictable and consistent performance. AVL Trees are simpler to understand compared to Red-Black Trees. For more details, visit the GeeksforGeeks article: <a href="https://www.geeksforgeeks.org/introduction-to-avl-tree/">AVL Tree Data Structure</a>.

AVL Tree Data Structure

We have discussed AVL insertion in the previous post. In this post, we will follow a similar approach for deletion.Steps to follow for deletion.&nbsp; To make sure that the given tree remains AVL after every deletion, we must augment the standard BST delete operation to perform some re-balancing. Following are two basic operations that can be performed to re-balance a BST without violating the BST property (keys(left) &lt; key(root) &lt; keys(right)).&nbsp; 1) Left Rotation&nbsp; 2) Right RotationAVL Tree - Deletion : <a href="https://www.geeksforgeeks.org/avl-tree-set-2-deletion/">https://www.geeksforgeeks.org/avl-tree-set-2-deletion/</a>

AVL Tree - Deletion

Segment Tree enables efficient range queries and updates on array intervals. It uses a divide and conquer approach, dividing the array into two parts recursively until single elements are reached. The tree is represented as an array with children of node i at (2*i + 1) and (2*i + 2). Operations like sum, max, GCD, XOR over ranges are performed in O(Log n). It supports dynamic updates while maintaining query efficiency.For more details, visit the GeeksforGeeks article: <a href="https://www.geeksforgeeks.org/introduction-to-segment-trees-2/">Introduction to Segment Trees</a>.

Introduction of Segment Tree

We have discussed the level order traversal of a tree. The idea is to print the last level first, then the second last level, and so on. Like Level order traversal, every level is printed from left to right.We can easily modify the method 1 of the normal level order traversal. In method 1, we have a method printGivenLevel() which prints a given level number. The only thing we need to change is, instead of calling printGivenLevel() from the first level to the last level, we call it from the last level to the first level.&nbsp; &nbsp;Reverse Level Order Traversal: <a href="https://www.geeksforgeeks.org/reverse-level-order-traversal/">https://www.geeksforgeeks.org/reverse-level-order-traversal/</a>

Reverse Level Order Traversal

Given a Binary Search Tree (BST) and a range [min, max], remove all keys which are outside the given range. The modified tree should also be BST. For example, consider the following BST and range [-10, 13].&nbsp; There are two possible cases for every node.Node’s key is outside the given range. This case has two sub-cases.&nbsp; Node’s key is smaller than the min value.&nbsp; Node’s key is greater than the max value.Remove BST keys outside the given range &nbsp;: <a href="https://www.geeksforgeeks.org/remove-bst-keys-outside-the-given-range/">https://www.geeksforgeeks.org/remove-bst-keys-outside-the-given-range/</a>

Remove BST keys outside the given range

Write a function to detect if two trees are isomorphic. Two trees are called isomorphic if one of them can be obtained from other by a series of flips, i.e. by swapping left and right children of a number of nodes. Any number of nodes at any level can have their children swapped. Two empty trees are isomorphic.We simultaneously traverse both trees. Let the current internal nodes of two trees being traversed be n1 and n2 respectively. There are following two conditions for subtrees rooted with n1 and n2 to be isomorphic.&nbsp;<ol><li>Data of n1 and n2 is same.</li><li>One of the following two is true for children of n1 and n2<ul><li>……a) Left child of n1 is isomorphic to left child of n2 and right child of n1 is isomorphic to right child of n2.</li><li>……b) Left child of n1 is isomorphic to right child of n2 and right child of n1 is isomorphic to left child of n2. &nbsp;</li></ul></li></ol>Tree Isomorphism Problem: <a href="https://www.geeksforgeeks.org/tree-isomorphism-problem/">https://www.geeksforgeeks.org/tree-isomorphism-problem/</a>

Tree Isomorphism Problem

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