No. Both child subtrees may contain values above 2, and either could contain 17. Heap ordering exposes a minimum; it does not divide all values into a smaller side and a larger side.

In a strict BST, every value in the left subtree is smaller than the node, and every value in the right subtree is larger. This is stronger than checking only immediate children. With search-tree height h, lookup takes O(h). Balancing can keep h=O(log n); ordinary insertion does not guarantee balance.

A binary heap maintains a complete tree shape, commonly stored in a list. In a min-heap, the smallest value is at index zero. Repeatedly selecting the cheapest pending task is exactly its strength.

from heapq import heappop, heappush

pending = []
for cost in [8, 2, 7, 1]:
    heappush(pending, cost)
assert pending[0] == 1
assert [heappop(pending) for _ in range(4)] == [1, 2, 7, 8]

Push and pop cost O(log n); reading the minimum is O(1). A heap’s list is not fully sorted. Finding an arbitrary value can require O(n) inspection.

Need repeated exact membership with hashable keys? A set is often the simpler fit, with expected constant-time operations under usual hashing assumptions. Need prefix completions? A trie preserves a different relationship again.

Analogy limit: “tree” specifies relationships, not an operation’s speed. Ordering rules, shape guarantees, and representation determine what work can be avoided.

No. Sorting once directly answers that workload. A heap becomes especially useful when selection and insertion are interleaved.