Trees and Binary Search Trees Questions

Approach
Inorder traversal means: visit the left child, then the node itself, then the right child. I would use recursion because it matches the tree structure naturally.

def inorder_traversal(root):
    result = []

    def dfs(node):
        if node is None:
            return
        dfs(node.left)
        result.append(node.val)
        dfs(node.right)

    dfs(root)
    return result

Worked example
For a tree with root 2, left child 1, and right child 3, the result is [1, 2, 3].

Empty tree
If root is None, the helper returns immediately and the answer is an empty list. That is the correct result because there are no nodes to visit.

Why recursion can fail
Python has a finite call stack. On a pathologically deep tree, such as a chain of 10,000 nodes, recursive calls can raise RecursionError even though the logic is correct. In that case, I would switch to an iterative traversal with an explicit stack.

What I would change before shipping
For production, I would remove recursion from any traversal or path algorithm that could see a deep tree. An explicit stack is safer than trying to raise Python's recursion limit, because sys.setrecursionlimit only hides the risk and can still crash the process on very deep inputs.

Concrete changes

• Replace recursive traversals with iterative versions.
• Use explicit stacks for inorder, postorder, and path-sum checks.
• Add tests for degenerate trees, such as a single chain of 20,000 nodes.
• Decide what to do on missing nodes or empty trees, and return that consistently.
• Add observability, such as logging maximum depth seen in production.

Example
If a customer uploads a tree shaped like a linked list, a recursive path-sum solution may fail even though the logic is correct. An iterative DFS with (node, state) pairs avoids that failure mode.

Production rule
My default is: if input depth is unbounded, do not rely on recursion. Use iterative code and make memory use predictable.

Approach
Use recursion with backtracking. A path sum is the sum of values from the root to a leaf. Because node values can be negative, I would not prune when the running sum gets too large, since later negative values could bring it back down.

def has_path_sum(root, target_sum):
    def dfs(node, remaining):
        if node is None:
            return False

        remaining -= node.val
        if node.left is None and node.right is None:
            return remaining == 0

        return dfs(node.left, remaining) or dfs(node.right, remaining)

    return dfs(root, target_sum)

Worked example
For the path 5 -> 4 -> 11 and target 20, the recursion checks 20 - 5 = 15, then 15 - 4 = 11, then 11 - 11 = 0, so it returns True.

Why backtracking is enough
Each recursive call gets its own remaining value, so there is no shared state to clean up. The function explores one root-to-leaf path at a time and stops as soon as it finds a match.

Complexity
Time is O(n) in the worst case, and space is O(h) for the recursion stack, where h is tree height.

Approach
I would keep one mutable list for the current path, append when I go down a node, and pop when I backtrack. At a leaf, I join a copy of the current path into a string. That avoids corrupting earlier paths.

def binary_tree_paths(root):
    result = []
    path = []

    def dfs(node):
        if node is None:
            return

        path.append(str(node.val))

        if node.left is None and node.right is None:
            result.append('->'.join(path))
        else:
            dfs(node.left)
            dfs(node.right)

        path.pop()

    dfs(root)
    return result

Worked example
For a tree 1 -> 2 -> 5 and 1 -> 3, the output is ['1->2->5', '1->3'].

Memory cost
The extra working memory is O(h) for the current path, where h is tree height. But the returned output can be much larger, because the number of root-to-leaf paths can grow quickly with the number of leaves. If callers do not need all paths at once, a generator would be more memory-friendly.

Approach
This is postorder traversal, which means left child, right child, then node. Since recursion is off the table, I would use an explicit stack and a last_visited pointer to know when a node's right subtree has already been processed.

def postorder_traversal(root):
    result = []
    stack = []
    current = root
    last_visited = None

    while current is not None or stack:
        if current is not None:
            stack.append(current)
            current = current.left
        else:
            peek = stack[-1]
            if peek.right is not None and last_visited is not peek.right:
                current = peek.right
            else:
                result.append(peek.val)
                last_visited = stack.pop()

    return result

Stack walk-through
For root 1 with left 2 and right 3:

• Push 1, then push 2.
• Node 2 has no children, so pop and output 2.
• Back at 1, move to 3, push 3, then pop and output 3.
• Finally pop and output 1.

Why this works
The stack always holds the path to the current node. last_visited prevents revisiting a right subtree after it has already been processed. Time is O(n), space is O(h), where h is tree height.