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Data Structure1 Practice8 Interview

Binary Tree

Tree whose nodes have at most two children, enabling standard recursive traversal patterns. This guide turns that contract into deliberate Python operations, interview invariants, and a runnable practice loop.

Recognize it when

Consider Binary Tree when the prompt's constraints and required operations match this shape: Tree whose nodes have at most two children, enabling standard recursive traversal patterns.

Pybit studies a professional Binary Tree interview workspace with precise technical objects.
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Binary Tree Code Labs

Mental Model

A binary-tree node has at most left and right children. Each child begins an independent subproblem, so a recursive function should define exactly what it returns for one subtree. The empty child is a real base case, not an exception to patch after the recursion.

Node Shape and Base Cases

Choose whether height counts nodes or edges and state the empty-tree value before coding. With node-count height, an empty subtree has height zero and a leaf has height one. A complete-index array is useful for Labs, but None under an unreachable parent does not create a floating node.

Compute Bottom-Up Properties

Postorder recursion receives completed left and right results before computing the current node. Count is 1 + left_count + right_count; height is 1 + max(left_height, right_height); a leaf has no reachable children. The same pattern powers balance, diameter, and subtree aggregation.

Aggregate a Binary Tree Profile

Reference
def binary_tree_profile(level_order):
    def visit(index):
        if index >= len(level_order) or level_order[index] is None:
            return 0, 0, 0
        left = visit(2 * index + 1)
        right = visit(2 * index + 2)
        count = 1 + left[0] + right[0]
        height = 1 + max(left[1], right[1])
        leaves = 1 if left[0] == right[0] == 0 else left[2] + right[2]
        return count, height, leaves
    return list(visit(0))
Practice

Implement binary_tree_profile(level_order). Count reachable nodes, node-count height, and leaves in a complete-index array with None gaps.

Public tests

  • Verify reachable subtree aggregation

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Traverse Level by Level

A queue groups nodes by distance from the root. Capture the current queue size, remove exactly that many nodes, and enqueue their reachable children for the next level. This differs from DFS because the frontier owns a whole breadth boundary.

Collect Reachable Tree Levels

Reference
from collections import deque

def binary_tree_levels(level_order):
    if not level_order or level_order[0] is None:
        return []
    queue = deque([0])
    levels = []
    while queue:
        level = []
        for _ in range(len(queue)):
            index = queue.popleft()
            level.append(level_order[index])
            for child in (2 * index + 1, 2 * index + 2):
                if child < len(level_order) and level_order[child] is not None:
                    queue.append(child)
        levels.append(level)
    return levels
Practice

Implement binary_tree_levels(level_order). Return reachable values grouped by depth, ignoring entries below missing parents.

Public tests

  • Verify level boundaries and reachability

Loading interactive editor.If it does not appear, the reference and starter code remain readable.Reload editor

Choose Recursive or Iterative State

Use recursion when the return value naturally summarizes a subtree and height is safe for Python’s call stack. Use an explicit stack for deep trees or controlled preorder/inorder traversal, and a queue for level order or minimum-edge distance. The Python recursion reference(opens in a new tab) does not guarantee tail-call elimination.

Common Pitfalls

  • Mixing edge-count and node-count height creates off-by-one answers.
  • Treating every non-None array entry as reachable accepts children of missing parents.
  • Recomputing subtree height inside every balance check can produce O(n squared) work.
  • Sharing a mutable result list across recursive calls leaks state between branches or test cases.

Explain It in an Interview

Say what the helper returns for one subtree and verify the empty node first. Then show how left and right results combine without revisiting nodes. State O(n) time because each reachable node is processed once; auxiliary space is O(h) recursion or O(w) queue width, depending on traversal.

Profile a Binary Tree(opens in a new tab) practices aggregation. Binary Tree Right Side View(opens in a new tab) requires choosing breadth order and the visible node from each level.

Python Version Note

Python 3.11+

The examples use standard Python containers and syntax available in the supported browser runtime; Binary Tree itself is taught as an interview abstraction.

Verify in Python docs(opens in a new tab)

Complexity & Invariants

Interview bounds depend on the stated representation, input model, and real Python operations.

OperationAverageWorstInterview note
Core Binary Tree workflowO(n)O(n)Breadth-first traverse reachable complete-tree indexes while tracking each node level and whether it has present children.

Space

O(n) for the demonstrated Binary Tree workflow.

Assumptions

  • State the concrete operation and representation before claiming a bound; tree height and graph density can change it.
  • The bound counts the operations in Profile a Binary Tree and does not hide Python slicing, sorting, or copying.

Invariants Worth Saying Aloud

  1. State the precise Binary Tree invariant before coding and preserve it after every update, traversal step, or recursive return.
  2. Traverse only nodes reachable from the root. This remains true after every accepted operation.
  3. Distinguish total nodes, root-to-leaf height, and leaf count. This remains true after every accepted operation.

When To Use Or Avoid Binary Tree

Use It When

  • Use Binary Tree when its representation directly supports the prompt's repeated query or update.
  • Use it when the invariant can be stated before coding and preserved after every operation.

Choose Another Tool When

  • Avoid it when a simpler Python sequence or mapping communicates the same contract with less state.
  • Avoid it when building the structure costs more time or space than the number of required queries can justify.

Common Pitfalls

These are the mistakes most likely to survive a happy-path example and fail a boundary case.

Leaving the core transition unfinished

Placeholder code cannot preserve the Binary Tree invariant or satisfy the public contract.

Prevent it: Implement the smallest complete transition and run the public tests before adding optimizations.

Avoid

def binary_tree_profile(level_order):
    pass

Use instead

from collections import deque

def binary_tree_profile(level_order):
    if not level_order or level_order[0] is None:
        return [0, 0, 0]
    queue = deque([(0, 1)])
    count = 0
    height = 0
    leaves = 0
    while queue:
        index, level = queue.popleft()
        count += 1
        height = max(height, level)
        children = [child for child in (2 * index + 1, 2 * index + 2) if child < len(level_order) and level_order[child] is not None]
        if not children:
            leaves += 1
        for child in children:
            queue.append((child, level + 1))
    return [count, height, leaves]

Where you will hit this: Profile a Binary Tree(opens in a new tab)

Breaking the central invariant

Counts every non-None array entry, including values that are unreachable below a missing parent.

Prevent it: Keep this invariant visible while editing: State the precise Binary Tree invariant before coding and preserve it after every update, traversal step, or recursive return.

Avoid

def binary_tree_profile(level_order):
    values = [value for value in level_order if value is not None]
    return [len(values), len(values), 1 if values else 0]

Use instead

from collections import deque

def binary_tree_profile(level_order):
    if not level_order or level_order[0] is None:
        return [0, 0, 0]
    queue = deque([(0, 1)])
    count = 0
    height = 0
    leaves = 0
    while queue:
        index, level = queue.popleft()
        count += 1
        height = max(height, level)
        children = [child for child in (2 * index + 1, 2 * index + 2) if child < len(level_order) and level_order[child] is not None]
        if not children:
            leaves += 1
        for child in children:
            queue.append((child, level + 1))
    return [count, height, leaves]

Where you will hit this: Profile a Binary Tree(opens in a new tab)

Hiding Python work inside the loop

Handle empty state, duplicate values, aliasing, and serialization boundaries explicitly in tests.

Prevent it: Account for every slice, copy, sort, membership test, and container mutation before claiming O(n).

Avoid

def binary_tree_profile(level_order):
    values = [value for value in level_order if value is not None]
    return [len(values), len(values), 1 if values else 0]

Use instead

from collections import deque

def binary_tree_profile(level_order):
    if not level_order or level_order[0] is None:
        return [0, 0, 0]
    queue = deque([(0, 1)])
    count = 0
    height = 0
    leaves = 0
    while queue:
        index, level = queue.popleft()
        count += 1
        height = max(height, level)
        children = [child for child in (2 * index + 1, 2 * index + 2) if child < len(level_order) and level_order[child] is not None]
        if not children:
            leaves += 1
        for child in children:
            queue.append((child, level + 1))
    return [count, height, leaves]

Where you will hit this: Binary Tree Right Side View(opens in a new tab)

Reviewed References

Official Python documentation supports language behavior. Canonical problem pages provide additional practice context.