construct bst from sorted array

If array has dupes then mid has to adjusted to rightmost dupe as left subtree <= node < right subtree for bst.

0123456
s  m  e
sme

01234
s m e
se
m

111
s e
 -m

11
se
-m

node atobst(a):
    if a == null:
        throw
    return atobst(a, 0, a.length()-1)

// a is asc
node atobst(a, start, end):
    if start > end:
        return null
    mid = start + (end-start)/2
    while mid+1 <= end && a[mid] == a[mid+1]:
        mid++
    n = new node(a[mid])
    n.left = atobst(a, start, mid-1)
    n.right = atobst(a, mid+1, end)
    return n

// a is desc
node atobst(a, start, end):
    if start > end:
        return null
    mid = start + (end-start)/2
    while mid-1 >= start && a[mid] == a[mid-1]:
        mid--
    n = new node(a[mid])
    n.left = atobst(a, mid+1, end)
    n.right = atobst(a, start, mid-1)
    return n

[Hat tip to MA]