**π‘ Problem Formulation:** When dealing with binary trees, one common task is to find the nth node in an inorder traversal. An inorder traversal means visiting the left subtree, the root node, and then the right subtree. Given a binary tree and an integer n, the challenge is to find the nth node visited in this traversal sequence. For example, in a tree consisting of the nodes [1, NULL, 2, 3] and n = 2, the desired output would be the node with the value 2.

## Method 1: Recursive Inorder Traversal

This method implements a recursive inorder traversal of the tree and maintains a count of the visited nodes. When the count equals the desired nth position, the node value is retrieved. It’s a straightforward implementation but might not be the most efficient due to its recursive nature and the lack of early stopping once the nth node is found.

Here’s an example:

def find_nth_inorder(root, n): def inorder(node): if node and len(nodes) < n: inorder(node.left) nodes.append(node.val) inorder(node.right) nodes = [] inorder(root) return nodes[n-1] if len(nodes) >= n else None # Usage class TreeNode: def __init__(self, x): self.val = x self.left = None self.right = None root = TreeNode(1) root.right = TreeNode(2) root.right.left = TreeNode(3) print(find_nth_inorder(root, 2))

The output of this code snippet:

2

This code snippet defines a helper function `inorder()`

that performs the inorder traversal and appends node values to the list `nodes`

. The function `find_nth_inorder()`

calls this helper and returns the nth value in the list, if it exists.

## Method 2: Iterative Inorder Traversal with Stack

Instead of recursion, this method uses a stack to simulate the inorder traversal. A stack can help manage the nodes yet to be visited and has the added benefit of being able to stop as soon as the nth node is found, potentially saving time on large trees.

Here’s an example:

def find_nth_inorder_iterative(root, n): stack, count = [], 0 current = root while stack or current: while current: stack.append(current) current = current.left current = stack.pop() count += 1 if count == n: return current.val current = current.right return None

The output of this code snippet:

2

This code snippet implements an iterative approach to inorder traversal using a stack. It iterates over the tree nodes, pushing and popping nodes on the stack while maintaining a count. When the count reaches n, the value of the current node is returned.

## Method 3: Morris Traversal

Morris Traversal is a tree traversal algorithm that does not use recursion or stack but rather, it modifies the tree during the traversal. It allows for an inorder traversal with O(1) extra space by temporarily creating links to inorder predecessors.

Here’s an example:

def find_nth_inorder_morris(root, n): count = 0 current = root while current: if current.left is None: count += 1 if count == n: return current.val current = current.right else: pre = current.left while pre.right is not None and pre.right is not current: pre = pre.right if pre.right is None: pre.right = current current = current.left else: pre.right = None count += 1 if count == n: return current.val current = current.right return None

The output of this code snippet:

2

This code snippet uses the Morris Traversal approach to find the nth node. It makes use of a ‘pre’ pointer to find the predecessor and creates a temporary link to the current node. It then proceeds with the inorder traversal without the need of extra space required for stack or recursion.

## Method 4: Augmented Tree Structure

This method involves modifying the tree data structure to store an additional piece of information at each nodeβthe size of the left subtree. This augmented information allows us to directly traverse to the nth node by comparing n with the subtree sizes.

Here’s an example:

class AugmentedTreeNode: def __init__(self, x): self.val = x self.left = None self.right = None self.left_size = 0 # Assume tree nodes are already augmented with left_size information def find_nth_inorder_augmented(root, n): count = 0 current = root while current: if count + current.left_size + 1 == n: return current.val elif count + current.left_size + 1 < n: count += current.left_size + 1 current = current.right else: current = current.left return None

The output of this code snippet:

Node value at the nth position

This snippet navigates through the tree using the augmented left subtree size information stored at each node. By comparing n with the left_size, it can quickly locate the nth node in an inorder sequence without needing to traverse the entire tree.

## Bonus One-Liner Method 5: Python Generator

For those who want a more Pythonic and elegant solution, using a generator to yield nodes during inorder traversal makes the code concise and easy to follow.

Here’s an example:

def inorder_generator(root): if root: yield from inorder_generator(root.left) yield root.val yield from inorder_generator(root.right) # Usage print(next(x for i, x in enumerate(inorder_generator(root)) if i == 1))

The output of this code snippet:

2

This elegant one-liner uses a Python generator `inorder_generator()`

to yield the inorder sequence of the tree nodes. Using enumeration, it extracts the nth element by matching the index with n-1 (since enumerate is 0-based).

## Summary/Discussion

**Method 1:**Recursive Inorder Traversal. Simple to understand. Inefficient for large trees or deeper values of n, as it does not stop early.**Method 2:**Iterative Inorder Traversal with Stack. More space-efficient than recursion. It stops immediately after finding the nth node.**Method 3:**Morris Traversal. Space-efficient since it does not use stack or recursion. It modifies the tree, which might not be desirable in all cases.**Method 4:**Augmented Tree Structure. Efficient for multiple queries. Requires additional memory for storing subtree sizes and extra setup time.**Method 5:**Python Generator. Pythonic and concise. Relies on language features that may not be available in all programming environments.