Trees and Binary Search Trees Questions

Design
I would compare nodes by identity, not by value, because duplicate values make value-based matching ambiguous. To avoid false positives, the helper should return both the candidate ancestor and whether each target was actually found.

def lowest_common_ancestor(root, p, q):
    def dfs(node):
        if node is None:
            return None, False, False

        left_lca, left_has_p, left_has_q = dfs(node.left)
        if left_lca is not None and left_has_p and left_has_q:
            return left_lca, True, True

        right_lca, right_has_p, right_has_q = dfs(node.right)
        if right_lca is not None and right_has_p and right_has_q:
            return right_lca, True, True

        has_p = left_has_p or right_has_p or node is p
        has_q = left_has_q or right_has_q or node is q

        if has_p and has_q:
            return node, True, True

        return None, has_p, has_q

    lca, has_p, has_q = dfs(root)
    return lca if has_p and has_q else None

Worked example
If p is in the left subtree and q is in the right subtree, the root becomes the LCA. If q is missing from the tree, the final has_q flag stays false, so the function returns None instead of a misleading ancestor.

Why this is safe
The function only reports an ancestor when both targets are confirmed present in the tree.

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.

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.

Conventions first
I would define height as the number of edges on the longest path from a node to a leaf. That makes an empty tree height -1 and a leaf height 0. I would define depth as the number of edges from the root to a node, so the root has depth 0. That keeps the API consistent.

def tree_height(root):
    if root is None:
        return -1
    return 1 + max(tree_height(root.left), tree_height(root.right))


def node_depth(root, target):
    def dfs(node, depth):
        if node is None:
            return None
        if node is target:
            return depth
        left = dfs(node.left, depth + 1)
        if left is not None:
            return left
        return dfs(node.right, depth + 1)

    return dfs(root, 0)

Worked example
If the tree is 10 -> 4 -> 2, then tree_height(root) is 2, node_depth(root, 10) is 0, and node_depth(root, 2) is 2.

API design choice
I would return None if the target node is not present, rather than guessing. That makes the behavior explicit and avoids ambiguity.

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.