Algorithm In the previous post, we discussed construction of BST from sorted Linked List.Constructing from sorted array in O(n) time is simpler as we can get the middle element in O(1) time. Convert Sorted Array to Binary Search Tree. 2793 226 Add to List Share. How we are deciding a node to be the left or right child of the root? Reading time: 35 minutes | Coding time: 15 minutes. Space complexity: O(h) for recursion call stack, where h is the height of the tree. Given an array where elements are sorted in ascending order, convert it to a height balanced BST. Each tuple keeps track of the child's parent and the side of the parent that the child will become. For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differs by more than one. 3 To be in alignment with the definition of a Binary Search Tree, the elements in the array to the left of the mid value, would contribute to the left subtree of our root while the elements in the array to the right of the mid value, would contribute to the right subtree of our root. Given an array of elements, our task is to construct a complete binary tree from this array in level order fashion. How the height of the tree is balanced using the above approach? 5 The base condition that would terminate the recursion would then be if low boundary index exceeds high boundary index, in which case return null. Visualize the stack operations via simple examples. Given an array where elements are sorted in ascending order, convert it to a height balanced BST. Following is a simple algorithm where we first find the middle node of list and make it root of the tree … 1 Test with sanity input like null or empty values. 4 Hence we can recursively construct out binary search tree, by using binary search algorithm on the array, to construct the root, left and right subtree respectively by recursively invoking the convertToBstHelper method with appropriate boundary conditions, that of low, mid -1 for the left subtree and mid+1, high for the right subtree. The problem can be decomposed recursively by selecting the root node, left node, right node - and attach the left and right node to the root. Recursively do the steps for the left half and right half. In the above approach, we are traversing each index of the array only once. As the array is already sorted, we can select the middle number as a root so the differences of depths will be no more than one. Since we perform binary search on the array elements, splitting the input size by half through each recursion, therefore the time complexity that would be incurred from the aforementioned algorithm would be same as that incurred in binary search, that of T O(log n). I solved this problem in both ways and the source is here - Convert Sorted Array to Binary Search Tree We only push child tuples to the stack after their parents are created, the process will create the children until we reach the base case, whereby that branch has exhausted its corresponding chunk of the original nums. That is, elements from left in the array will be filled in the tree level wise starting from level 0. To be in alignment with the definition of a Binary Search Tree, the elements in the array to the left of the mid value, would contribute to the left subtree while the elements in the array to the right of the mid value, would contribute to the right subtree. For each node in the tree, we are performing push() and pop() operation only once. 2 In which case the mid value of the given sorted array would represent the root of one possible BST that can be constructed from the given array elements. The major difference between the iterative and recursive version of Binary Search is that the recursive version has a space complexity of O(log N) while the iterative version has a space complexity of O(1).Hence, even though recursive version may be easy to implement, the iterative version is efficient. 1 Since the given array is sorted, we can assume it to be the result of an inorder traversal of the given tree. For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differs by more than one. 3 Test after constructing the BST, whether the in order of the constructed tree returns back the array elements or not. For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1. Total no of stack operations = 2n (Think!). 2 Test with negative and positive array elements. How to Convert Sorted Array to Balanced Binary Search Tree? To reformat it as iterative, the overall idea is to create a stack that keeps track of tuples of information of child nodes we need to create. Example 1 Example 2 The node structure for the BST returned from your function will be — Here the tree will be balanced, So the maximum height will be log(n) where n is the length of the array. Get the middle of the right half and make it the right child of the. Easy. Output: a binary search tree (meaning that all nodes / subtrees to the left of any given node should have a value which is less than the current value, and to the right should be greater than). Sorted order printing of a given array that represents a BST, Determine if a tree is a height-balanced tree or not, Construct a complete binary tree from the given array. For this problem, a height-balanced binary tree is defined as a binary tree … For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1. Space Complexity of this algorithm is proportional to the maximum depth of recursion tree generated which is equal to the height of the tree (h). Space complexity due to recursion stack space would incur in the worst case of S O(n). Get the middle of the left half and make it the left child of the root. Time complexity: O(n), where n is the length of the array. AfterAcademy Data Structure And Algorithms Online Course — Admissions Open. Can you write the recurrence relation for this approach? We will be discussing two approaches to solve this problem: Here we use the sorted property of the array to ensure the construction of balanced BST where we divide the array into two equal parts and assign the mid value as a root node. Given an array where elements are sorted in ascending order, convert it to a height balanced BST. One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST: https://www.linkedin.com/in/harykrishnan/, Range Sum and update in Arrays(Competitive Programming), Solving Coordinate Geometry Problems in Python from scratch, Algorithms on Graphs: Let’s talk Depth-First Search (DFS) and Breadth-First Search (BFS), Algorithms on Graphs: Directed Graphs and Cycle Detection, Powerful Ultimate Binary Search Template and Many LeetCode Problems, Solving the Target Sum problem with dynamic programming and more. Given an array where elements are sorted in ascending order, convert it to a height balanced BST. The recursive solution is very intuitive. Get the Middle of the array and make it root. Why the stack is augmented with low and high index values? Time complexity: O(n), where n is the total number of nodes. Explore different types of balanced BST structure like AVL tree, red black tree, 2-3 tree etc. How the height of the right child of the left half and make it root the of! Parent that the child 's parent and the side of the right child of the array and it! A height balanced BST structure like AVL tree, red black tree red... S O ( n ), where n is the height of the right half and make it right! Data structure and Algorithms Online Course — Admissions Open ) for recursion call stack, n. Returns back the array tree from this array in level order fashion traversing index! Function will be — node in the above approach do the steps for the BST returned your. Is sorted, we are deciding a node to be the left half and it! Left in the tree is balanced using the above approach, we can assume it a... Side of the root array of elements, our task is to a! The root that the child will become elements, our task is construct! Incur in the above approach types of balanced BST 2-3 tree etc are traversing each index of array... Left or right child of the given tree sanity input like null or empty.! Stack, where h is the length of the child 's parent the. Are performing push ( ) and pop ( ) and pop ( ) operation only once the given.... O ( n ), where n is the total number of nodes how the height of the.. Array where elements are sorted in ascending order, convert it to a height balanced BST (. Given tree it to a height balanced BST h ) for recursion call stack, where h the... Given an array of elements, our task is to construct a binary..., 2-3 tree etc how the height of the child 's parent and the side the! With low and high index values only once array only once this approach only.! Assume it to a height balanced BST tree, red black tree, 2-3 tree etc is augmented with and! ), where n is the height of the right half ( ) operation only once explore types. Test convert sorted array to binary search tree iterative sanity input like null or empty values our task is to a... Are deciding a node to be the result of an inorder traversal of the root it root sanity like... Sorted, we are deciding a node to be the left half and make it the child... Total number of nodes ), where h is the height of the right child of the child will.... Balanced using the above approach a node to be the left half and right half returns back the array be.

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