get 2nd largest item in bst

(n):
    if n == null:
        return
    // max2 could be parent of max
    max2 = null
    max = n
    while max.right != null:
        max2 = max
        max = max.right
    // if max.left is not null then max2 is
    // max.left's rightmost
    if max.left != null:
        max2 = max.left
        while max2.right != null:
            max2 = max2.right
    return max2

Using a minheap of capacity k will work for any k (k = 2+).

// example 1
       4
   3       6 max2
 2   4    5  7
1 3

// example 2
       4
   3
 2   4 max2
1 3

With duplicates and minheap(2, n => n.value), traversing the left subtree will give wrong node, node 1 of level 2 (node 1 of level 1 with be displaced by node 2). Traversing right subtree first (as there are no dupes) will fix this but will not work for 3rd largest. To fix that minheap should look at value and if same then at level (same value nodes with greater level get pushed up), so minheap(k, comparer).

// dupes example
  1
1   2

// space: O(k), time: O(nlogk)
(n, k, minheap = minheap(2, comparer)):
    if n == null:
        return
    if minheap.count() == k: // minheap.capacity() is k
        if n.value > minheap.peek().value:
            minheap.displace(n)
    if minheap.count() < k:
        minheap.insert(n)
    (n.right, k, minheap)
    (n.left, k, minheap)

comparer:
    // node with greater value is greater
    // for same value, node with lesser level is greater
    int compare(n, other):
        c = n.value.compare(other.value)
        if c == 0:
            c = n.level.compare(other.level) * -1
        return c

[Hat tip to IC]