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  4. Trees: Is This a Binary Search Tree?
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Trees: Is This a Binary Search Tree?

  • Dan_Golding + 1 comment

    I ended up with pretty much the same code but instead of a closure variable I just passed the variable in. I used a single element list since it's gets passed by reference which allows changes to persist when going back down the call stack:

    def checkBST(root):
    return(check_in_order(root,[-1]))
    
def check_in_order(root,prev):
    result = True
    if root.left is not None:
        result &= check_in_order(root.left,prev)
    if prev[0] >= root.data:
        return False
    prev[0] = root.data
    if root.right is not None:
        result &= check_in_order(root.right,prev)
    return result
    • svdubl + 0 comments

      thank you! I started with the same elegant code as yours but before I figured out that you can pass by reference with the list I had to change it to this bulky code:

      def inOrderTraversalCheck(node, inValue=None):
	result = True
	if node.left is not None:
		(result, value) = inOrderTraversalCheck(node.left,min(node.data,inValue))
		if result is False:
			return (False, value)
		if value >= node.data:
			return (False, value)
	if node.right is not None:
		(result, value) = inOrderTraversalCheck(node.right, max(node.data,inValue))
		if result is False:
			return (False, value)
		if value <= node.data:
			return (False, value)
	if inValue is not None:
		if inValue >= node.data:
			return (False, node.data)
	if node.right or node.left is not None:
		return (True, max(node.data,value))
	else:
		return (True,node.data)


def checkBST(root):
	(result, value) = inOrderTraversalCheck (root)
	return (result)