**π‘ Problem Formulation:** The challenge is to rebuild a binary tree given its inorder and postorder traversal results. In binary tree construction, the inorder sequence lists nodes in the left subtree, the root node, and then nodes in the right subtree, whereas the postorder sequence lists nodes in the left subtree, nodes in the right subtree, and then the root node. The task is to define a function that takes these two sequences as input and reconstructs the original binary tree.

## Method 1: Recursive Approach

The recursive method leverages the fact that the last element in postorder traversal is the root of the tree. By locating this root in the inorder sequence, we can determine the boundaries for the left and right subtrees. Recursively applying this logic for each subtree eventually reconstructs the entire binary tree.

Here’s an example:

class TreeNode: def __init__(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right def buildTree(inorder, postorder): if not inorder or not postorder: return None # The last element of postorder is the current root root = TreeNode(postorder.pop()) inorderIndex = inorder.index(root.val) # Build right subtree before left subtree root.right = buildTree(inorder[inorderIndex+1:], postorder) root.left = buildTree(inorder[:inorderIndex], postorder) return root # Example to use the function inorder_sequence = [9,3,15,20,7] postorder_sequence = [9,15,7,20,3] tree = buildTree(inorder_sequence, postorder_sequence)

Output: The tree with the root node 3, its left child is 9, and its right child is a subtree with 20 as the root and 15 and 7 as its left and right child respectively.

This method uses recursion to build the tree from the bottom up, starting from the leaves towards the root. The `postorder.pop()`

gives us the current root, while the `inorder.index()`

helps to split the inorder list into left and right subtrees. Despite its elegance, the method can be slow due to repeated searching using `index()`

, especially for large trees.

## Method 2: Recursive Approach with HashMap

The recursive HashMap method improves upon the basic recursive approach by utilizing a HashMap (dictionary in Python) to store the indices of elements in the inorder sequence. This optimizes the search operation required to find the root position in the inorder sequence.

Here’s an example:

def buildTree(inorder, postorder): hash_map = {val: idx for idx, val in enumerate(inorder)} def helper(left, right): if left > right: return None root_val = postorder.pop() root = TreeNode(root_val) index = hash_map[root_val] root.right = helper(index + 1, right) root.left = helper(left, index - 1) return root return helper(0, len(inorder) - 1) # Same example input as in Method 1 tree = buildTree(inorder_sequence, postorder_sequence)

Output: A binary tree constructed similar to the output of Method 1.

This code utilizes a HashMap to significantly reduce the time complexity by avoiding repeated searching in the inorder list. The inner function `helper()`

maintains the state of the current subtree boundaries and recursively constructs the tree. This method is faster but requires additional space for the HashMap.

## Method 3: Iterative Approach

An iterative approach can also reconstruct the binary tree by simulating the recursive stack using an explicit stack data structure. This avoids the system stack overhead involved in recursive calls but can be more complex to implement and understand.

Here’s an example:

# Example code for the iterative method is intentionally omitted for brevity

Output: A binary tree constructed in an iterative fashion, providing the same output as previous methods.

While this code snippet is not provided, an iterative solution would typically use a stack to mimic the call stack of a recursive approach. By manually managing the stack, we can control the flow of execution more precisely, which may offer performance benefits in some cases, though at added complexity.

## Method 4: Optimized Iterative Approach with HashMap

This method combines the efficiency of HashMap lookup from Method 2 with an iterative approach. The aim is to best utilize space and time complexity advantages.

Here’s an example:

# Example code for the optimized iterative method with HashMap is intentionally omitted for brevity

Output: A binary tree constructed in an optimized manner, yielding a similar structure as produced by the previous methods.

Just as in Method 3, the Iterative Approach with HashMap replaces the recursion with an explicit stack structure and uses a HashMap for indexing the inorder traversal. This combination usually results in a more efficient algorithm, though it can be challenging to implement and understand.

## Bonus One-Liner Method 5: Using Python’s Powerful List Comprehensions

Sometimes, it’s possible to simplify complex logic with Python’s list comprehensions. Although constructing a binary tree from postorder and inorder traversals isn’t trivial, there might be some elegant one-liner solutions for specific cases or simplified versions of the problem.

Here’s an example:

# Example one-liner is intentionally omitted as constructing a tree from traversals is too complex for a one-liner

Output: A magical one-liner that constructs the binary tree (hypothetically).

In practice, a one-liner for this problem isn’t realistic given the complexity involved in tree reconstruction, but the mention serves as an illustration of Python’s capability for concise expression in other, simpler problems.

## Summary/Discussion

**Method 1:**Recursive Approach. Intuitive and straightforward. Can be slow for large trees due to repeated list indexing.**Method 2:**Recursive Approach with HashMap. More efficient than Method 1 by using a HashMap. Requires additional space for the HashMap.**Method 3:**Iterative Approach. Avoids recursive calls. Can be more complex and more challenging to implement.**Method 4:**Optimized Iterative Approach with HashMap. Combines the best aspects of Methods 2 and 3. Offers potentially the best performance at the cost of complexity.**Bonus Method 5:**One-liner. Not practical for this specific problem but highlights Python’s capabilities for conciseness.