Where does this level end?

The queue starts [A,B]. A adds C,D, and B adds E.

After removing A it is [B,C,D]. Which items still belong in today’s result?

Derive the one piece of state that records the boundary.

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Binary Tree Level Order Traversal — predict before coding

A queue orders nodes by distance from the root. Capturing its current length freezes one level before newly discovered children are added.

Decision: Why must the level loop use a saved size rather than a changing len(queue) while children are appended?

Presenter notes

Model answer: The queue length changes during processing. A saved size separates current-level nodes from children that belong to the next level.

Correctness argument: At the start of each outer iteration, the queue contains exactly the next level's nodes in left-to-right order. Freeze that count and remove exactly that many nodes. Appending each node's left child before its right child builds the next level in the same left-to-right order, because all children of an earlier parent precede children of a later parent. The frozen count prevents these new nodes from being consumed as part of the current level. Recording the removed values therefore gives one correct level, and induction processes all depths exactly once.

Debugging prompt: Test an empty tree and a sparse tree where parents have different child counts. A complete tree can hide mistakes that happen only at irregular level boundaries.

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Binary Tree Zigzag Level Order Traversal — predict before coding

Traversal order and output placement can be separated. Keep ordinary left-to-right BFS discovery, but write alternate levels into reversed result positions.

Decision: If a BFS level is [4,5,6,7] and this depth is reversed, where is the first removed value 4 written?

Presenter notes

Model answer: At index 3, producing the final level [7,6,5,4]. Discovery can remain left-to-right.

Correctness argument: BFS supplies each level's nodes in left-to-right order by the usual queue invariant. On a forward level, store the ith removed value at index i; on a reversed level, store it at size-1-i. Each mapping is a bijection on the level's indices, so every value appears exactly once in the required direction. Child enqueue order remains left then right regardless of the output direction, preserving the next level's discovery invariant. Toggling only the placement direction after each frozen-size level produces the required alternation without changing tree topology.

Debugging prompt: Do not reverse child insertion and output placement simultaneously unless you have proved the resulting order. Use at least three levels so the direction toggles twice.

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Binary Tree Right Side View — predict before coding

The visible node at a depth is the rightmost existing node there, which may come from a left subtree when the right subtree is shallower.

Decision: A root has left child 2 with right child 5, and right child 3 with no children. What is the right-side view?

Presenter notes

Model answer: [root value,3,5]. Node 5 is visible at the deeper level despite belonging to the root's left subtree.

Correctness argument: BFS processes nodes at each depth in left-to-right order. Therefore the final node among the frozen current-level count is precisely the rightmost existing node at that depth. Recording its value yields the visible value for that level. Appending children left before right preserves this ordering at the next depth. Repeating until the queue is empty visits all levels, including deeper levels where only a left-side branch survives. The algorithm does not assume that following root.Right repeatedly reaches every visible node.

Debugging prompt: Test a deeper left branch and missing right children. A routine that simply follows Right pointers is not a right-side-view algorithm.

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Average of Levels in Binary Tree — predict before coding

A level average requires a per-level accumulator and a frozen count. Integer division and cross-level sums are independent sources of subtle errors.

Decision: For a level containing values 2 and 3, what is the average and why is integer division wrong?

Presenter notes

Model answer: 2.5. Computing 5/2 with integer operands first loses the fractional part before a later conversion to float64.

Correctness argument: At each level boundary, freeze the number of nodes and reset the sum to zero. Processing exactly that many queued nodes adds each value from that depth once and no values from other depths. The arithmetic mean is therefore this sum divided by the frozen count, converted to floating point. Enqueuing children preserves the next-level order and coverage invariant but does not change the current divisor. Induction supplies the correct mean for every depth. For the stated 32-bit node values and at most 100000 nodes, an int64 or exact float64 integer accumulation is safe for each level sum.

Debugging prompt: Use unequal-width levels, negative values, and large 32-bit values. Compare floating results with an explicit tolerance; do not require a particular decimal string formatting.

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Populating Next Right Pointers in Each Node II — predict before coding

Sparse trees require linking the actual nodes present at each level, not assuming every parent has two children. Existing Next pointers are output fields and cannot be trusted as traversal input in the queue reference.

Decision: Why must the last node’s Next be set to nil even when a newly built tree would already have nil there?

Presenter notes

Model answer: Callers may provide stale Next pointers or invoke Connect again. Leaving the terminal field unchanged could retain a cross-level edge or cycle.

Correctness argument: BFS forms the exact left-to-right node sequence for one level. Link each previously removed node's Next to the current node, then set the final node's Next to nil. These assignments create precisely the successor relation within that level and no cross-level edge. Child discovery uses Left and Right pointers, so stale or cyclic initial Next values cannot corrupt traversal. Repeating for every frozen-size level sets every Next field to its required value, including terminal nils, while preserving the root, node identities, child edges, and values.

Debugging prompt: Prepopulate Next pointers with self-links or wrong-level pointers, then verify each exact level chain and its nil terminator. Also test gaps between parents' children.

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One frontier, several consumers

Root 6 has children 2,10. Node 2 has only right child 4; node 10 is a leaf.

Derive level order, zigzag, right view, averages, and Next links. Which results require every value, and which need only an aggregate?

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