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Binary Tree Iterative Inorder Traversal.       visit(node) For traversing a (non-empty) binary tree in in-order fashion, we must do these three things for every node N starting from root node of the tree: In normal in-order traversal we visit left subtree before the right subtree. In this post, let’s focus on the iterative implementation of inorder traversal or iterative inorder traversal without recursion.       s.push(node) As tree is not a linear data structure, from a given node there can be more than one possible next node, so some nodes must be deferred i.e stored in some way for later visiting. Implementation: Inorder traversal can be implemented in three ways – Recursively, Iteratively and using Morris traversal.We will discuss about the first two approaches in this article. We will recursively process the left subtree, then we will process the root node and then finally we will process the right subtree. For that purpose, we have traversals for different scenarios which can be really helpful.       node -> node.right. In this implementation, we are going to use a stack in place of recursion. When number of nodes in tree are less then we can go for recursive traversal but when we have millions of records then recursive traversal may give stackoverflow. There are basically three types of depth-first search algorithms in trees(or graphs) – Preorder, Postorder, and Inorder traversal. Do NOT follow this link or you will be banned from the site. We implemented those traversals in a recursive way. An inorder traversal on a BST returns a sorted list of node values. These operations can be defined recursively for each node. // else if current node is null, we pop an element from stack, // print it and finally set current node to its right child, # Iterative function to perform in-order traversal of the tree, # start from root node (set current node to root node), # if current node is None and stack is also empty, we're done, # if current node is not None, push it to the stack (defer it). Step 1: If the root is NULL i.e tree is empty, return.Step 2: Recursively process left subtree.Step 3: Process the root node and print its value.Step 4: Recursively process the right subtree. // Recursive function to perform in-order traversal of the tree, // Display the data part of the root (or current node), # Recursive function to perform in-order traversal of the tree, # Display the data part of the root (or current node), // Iterative function to perform in-order traversal of the tree, // start from root node (set current node to root node), // if current node is null and stack is also empty, we're done, // if current node is not null, push it to the stack (defer it). (40 votes, average: 4.38 out of 5)Loading... Guys please learn both the methods once in my interview the interviewer asked me to write both iterative and recursive approach . The idea of Morris Traversal is based on Threaded Binary Tree.In this traversal, we first create links to Inorder successor and print the data using these links, and finally revert the changes to restore original tree. Only i have written in both methods , my friends could only write recursive approach while i have written in both the methods and i got the job. I found this article very clear (like others that I’ve read from this website). Happy coding . In this post, let’s focus on the iterative implementation of inorder traversal or iterative inorder traversal without recursion. while (not s.isEmpty() or node != null) In Binary search tree traversals we discussed different types of traversals like inorder, preorder and postorder traversals. Thanks for sharing your concerns. s -> empty stack I'm using a stack for the traversal. Conflating the two meanings with a single symbol somehow blurs the mechanism a bit: Inorder traversing should give sorted nodes!! Problem Statement:  Give a binary tree, perform the inorder traversal and also print the elements. In an Inorder traversal, we process all the nodes of a tree by recursively processing the left subtree, then processing the root, and finally the right subtree. In this post, inorder tree traversal is discussed in detail. The inorder traversal on below tree: 40, 20, 10, 50, 30. Time Complexity: O(N) – In the inorder traversal, we visit each node of the tree exactly once, and, the work done per node (printing node value) is also constant – O(1), which makes the algorithm of O(N) time complexity. I hadn’t been satisfied with the way in which the iterative solution of inorder binary tree traversal has been explained so far in my searches on the intertubes. Tree traversal are methods to traverse tree in different ways. Use cases: Inorder traversal is mostly used with Binary Search Tree. Space Complexity: O(N) – If we have a skewed binary tree, then recursion has to go N nodes deep before it hits the end(or NULL or base case). References: https://en.wikipedia.org/wiki/Tree_traversal. Time Complexity: O(N) – In an Inorder Traverse, we traverse each node of the tree exactly once, and, the work done per node is constant i.e O(1) operation, hence the time complexity of an inorder traversal(recursive) is O(N). Here is my function for iterative inorder traversal. Approach 2 – Iterative implementation of Inorder Traversal. However, I believe the iterative algorithm would be clearer if it made a proper distinction between children nodes and popped nodes, as they have a different significance inside the loop. Approach 2 – Iterative implementation of Inorder Traversal. Write an iterative solution as a practice of depth first search. As we all know, a Binary Tree is a tree made up of nodes and each node can have at most two children. 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