/* ---------------------------------------------------------- */
/* | Extra functions | */
/* ---------------------------------------------------------- */
/// <summary>
/// Problem. Given the value of two nodes in a binary search tree,
/// find the lowest (nearest) common ancestor.
/// You may assume that both values already exist in the tree.
///
/// Solution. When your target values are both less than the current node, you go left.
/// When they are both greater, you go right.The first node you encounter
/// that is between your target values is the lowest common ancestor.
///
/// Complexity is O(log(n)). You travel down a path to the lowest common ancestor.
/// Recall that traveling a path to any one node takes O(log(n)).
/// </summary>
public Node <T> GetLowestCommonAncestor(T value1, T value2)
{
Node <T> root = this.Root;
while (root != null)
{
Interfaces.IComparable <T> value = root.Value;
if (value.IsGraterThan(value1) && value.IsGraterThan(value2))
{
root = root.Left;
}
else
{
if (value.IsLessThan(value1) && value.IsLessThan(value2))
{
root = root.Right;
}
else
{
return(root);
}
}
}
return(null); //only if empty tree
}
/* ---------------------------------------------------------- */
/* | Base functions | */
/* ---------------------------------------------------------- */
private Node <T> _InsertNode(Node <T> root, Interfaces.IComparable <T> value)
{
//place for adding was found
if (root == null)
{
root = new Node <T>(null, null, value);
}
else
if (value.IsLessThan(root.Value.Get()))
{
root.Left = _InsertNode(root.Left, value);
}
else
if (value.IsGraterThan(root.Value.Get()))
{
root.Right = _InsertNode(root.Right, value);
}
else
{
root.Inc();
}
return(root);
}