CareerCup

def find_node_with_value(root,node_value){
  stack = list()
  stack.push(root)
  while ( !empty(stack) ){
    node = stack.pop()
    if ( node_value == node.value ){
      return node 
    }
    if ( node.left != null ){
      stack.push(node.left)
    }  
    if ( node.right != null ){
      stack.push(node.right)
    }
  }
  return null
}
// i have an implicit root 
def find(x,y){
  x_node = find_node_with_value(root ,x)
  if ( x_node == null ) return false 
  y_node = find_node_with_value(x_node ,y)
  y_node != null // return this 
}

if y is contained in one of the 2 possible subtrees of x then x becomes LCA of x and y

Thank you for the reply, but this is not recursion.

Can you please restate the problem:
Do you want to know, if there is a subtree D in the tree T that has value x in the root of D and value y somewhere in D. If you write binary tree you assume no BST (no order)?

In this case something like:

def find(root, values, i):
  if root is None: return False
  if root.value == values[i]: i = i - 1
  if i < 0: return True
  return find(root.left, v, i) or find(root.right, v, i)

def find_if_descendant(root, x, y):
  return find(root, [x, y], 1)

@ChrisK thank you for the solution, this is sth more like that that I was looking for... maybe I took the question too literally.. maybe I could have used a recursive method in the background, and my initial method would have contained only 2 arguments.