In max heap, a small element may be present on either left or right we don't know hence we need to check left children as well as right children, in height of 2 max heap there can be max 4 leaf nodes and hence we need to check all of them. so height of 2 can contain 7 nodes out of which we need to check 4 leaf nodes to find smallest which is approximately equal to n/2..
If you notice the arrangement of array elements you fill find all leaf nodes are present in next half of the array. Hence you need scan n/2 elements to find smallest. actually it is floor(n/2)+1 where you can get position of the first leaf node.
In balanced binary search tree you can find minimum in logn time as we know for sure smallest will always be present on the left side. Hence time require to search element.
Where as in binary search tree it might take n comparisons to check for smallest element if its a skew tree.