[英]Remove from binary tree C#
嗨,我在从C#中的二叉树中删除时遇到了一个小问题。 我不知道为什么,但是这段代码无法正常工作(调用remove方法后我的树没有改变)。 这是我的代码:
public class BinaryTree<T>
{
public BinaryTree<T> Left, Right;
public T Data;
public BinaryTree()
{
this.Left = null;
this.Right = null;
}
}
public BinaryTree<T> FindNode(T value,ref BinaryTree<T> myTree)
{
if (myTree == null)
return null;
else
{
int result = Comparer<T>.Default.Compare(value, myTree.Data);
if (result == 0)
return myTree;
else if (result > 0)
return FindNode(value, ref myTree.Right);
else if (result < 0)
return FindNode(value, ref myTree.Left);
}
return myTree;
}
public void RemoveValue(T value,ref BinaryTree<T> myTree)
{
BinaryTree<T> helper = new BinaryTree<T>();
BinaryTree<T> MyTree = myTree;
if (MyTree == null) return;
MyTree =FindNode(value,ref MyTree);
if (MyTree.Left == null || MyTree.Right == null)
helper = MyTree;
else
{
helper = MyTree.Left;
while (helper.Right!=null)
helper = helper.Right;
MyTree.Data = helper.Data;
}
if (helper.Left == null)
helper = helper.Right;
else
helper = helper.Left;
}
BinaryTree表示树中的每个节点。
这是对您的代码的一些小整理,其中包含有关我认为RemoveValue
应该如何工作的建议。 我由您决定完成实施。
public class BinaryTree<T>
{
public T Data;
public BinaryTree<T> Left, Right;
public BinaryTree()
{
this.Left = null;
this.Right = null;
}
public static BinaryTree<T> FindNode(T value, BinaryTree<T> myTree)
{
if (myTree == null)
return null;
else
{
int result = Comparer<T>.Default.Compare(value, myTree.Data);
if (result == 0)
return myTree;
else if (result > 0)
return FindNode(value, myTree.Right);
else if (result < 0)
return FindNode(value, myTree.Left);
}
return myTree;
}
public static void RemoveValue(T value, ref BinaryTree<T> myTree)
{
if (myTree == null)
return;
BinaryTree<T> treeNodeToRemove = FindNode(value, myTree);
// First case: The node to remove has a single subtree
if(treeNodeToRemove.Left == null ^ treeNodeToRemove.Right == null)
{
// We need to change the Left||Right reference of our parent to us...
}
// Second case: Both subtrees are null
else if (treeNodeToRemove.Left == null && treeNodeToRemove.Right == null)
{
// We need to change the reference of our parent to null
}
// Third case: Both subtrees are full
else
{
// ...
}
}
}
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