PriorityQueue.iterator() — what order does it give you?

PriorityQueue.iterator() does not return elements in priority order. It walks the backing array, which is only heap-ordered, not sorted. The single guarantee is that peek() returns the head; everything after that is in whatever arrangement the heap happens to hold.

PriorityQueue<Integer> pq = new PriorityQueue<>();
pq.addAll(List.of(5, 1, 3, 2, 4));

for (int x : pq) System.out.print(x + " ");   // e.g. 1 2 3 5 4  -- NOT sorted
System.out.println();

while (!pq.isEmpty()) System.out.print(pq.poll() + " ");  // 1 2 3 4 5  -- sorted

Why

The heap invariant is local: each node ≤ its children. That orders root-to-leaf paths but says nothing about left-vs-right siblings or across subtrees. The array is a valid heap with many internal orderings, and the iterator simply exposes the raw array. The same applies to toString(), toArray(), stream(), and enhanced-for — all reflect array order.

How to actually get sorted order

• Drain it: repeatedly poll() until empty — each poll is O(log n), total O(n log n), and it destroys the queue.
• Copy and sort: pq.stream().sorted().toList() if you must keep the queue.
• Don't reach for iterator() expecting order — this is the classic trap.

The deeper interview point

This is the same lesson as "a hash set has no order": picking a structure for one fast operation (here, O(1) min) means giving up other guarantees (here, total order on iteration). If you need both fast-min and ordered iteration, a heap is the wrong tool — use a TreeSet/TreeMap.