Morris Traversal: Inorder Without Recursion Or Stack

Ever wondered how to traverse a binary tree in an inorder fashion without the usual suspects – recursion and a stack? It sounds like a bit of a puzzle, doesn't it? Well, the Morris Traversal algorithm is here to solve just that! It's a clever technique that allows you to visit nodes in inorder sequence, but it does so by temporarily modifying the tree's structure. This means you can achieve the traversal using only constant extra space, which is a huge advantage when dealing with massive trees where stack overflow or memory limitations could be a concern. The fundamental idea behind Morris Traversal is to find a way to return to the parent node after visiting its left subtree, without needing a stack to keep track of where to go next. It achieves this by utilizing the right child pointer of the predecessor node in the inorder sequence. This predecessor is the rightmost node in the left subtree of the current node. By making this predecessor's right child pointer point back to the current node, we create a temporary link. After visiting the left subtree and all its descendants, we can follow this temporary link back up to the current node, effectively resuming our traversal. Once we've processed the current node, we restore the original tree structure by setting the predecessor's right child pointer back to its previous state (usually null). This intricate dance of modifying and restoring pointers is what makes Morris Traversal so unique and memory-efficient. It’s a testament to elegant algorithmic design, solving a common problem with an uncommon approach.

Let's dive deeper into the mechanics of the Morris Traversal algorithm and understand how it achieves inorder traversal without relying on recursion or an explicit stack. The core challenge is navigating back up the tree after traversing the left subtree. In a typical inorder traversal using recursion, the call stack implicitly handles this. With an iterative approach using a stack, we explicitly push nodes onto the stack to remember where to return. Morris Traversal sidesteps this entirely by cleverly using the tree's right child pointers. The algorithm iterates through the tree, maintaining a current node pointer. For each current node, it first checks if it has a left child. If it doesn't, the node is simply printed (in inorder fashion), and we move to its right child. This is the easy part. The magic happens when a left child does exist. Here's where the predecessor comes into play. We find the inorder predecessor of the current node. The inorder predecessor is defined as the rightmost node in the current node's left subtree. To find it, we start at current->left and keep traversing right as far as possible. Once we find this predecessor, we check its right child pointer. If the predecessor's right child is null, it means we haven't visited this predecessor's subtree yet. So, we establish a temporary link: we set the predecessor's right child to point back to the current node. This link acts as our