二叉排序树的删除

yanglilibaobao 2007-04-18 10:41:36

二叉排序树的删除：
对于一般的二叉树来说，删去树中的一个结点是没有意义的，因为它将使以被删除的结点为根的子树变成森林，破坏了整棵树的结构
但是，对于二叉排序树，删去树上的一个结点相当于删去有序序列中的一个记录，只要在删除某个结点后不改变二叉排序树的特性即可。
在二叉排序树上删除一个结点的算法如下：
btree * DeleteBST(btree *b, ElemType x)
{
if (b)
{
if (b->data == x)
b = DelNode(b);
else if (b->data > x)
b->lchild = DeleteBST(b->lchild, x);
else
b->rchild = DeleteBST(b->rchild, x);
}
return b;
}

　　　其中删除过程有两种方法。
第一种过程如下：
1。若p有左子树，找到其左子树的最右边的叶子结点r，用该叶子结点r来替代p，把r的左孩子
作为r的父亲的右孩子。
2。若p没有左子树，直接用p的右孩子取代它。

第二种过程如下：
1。若p有左子树，用p的左孩子取代它；找到其左子树的最右边的叶子结点r，把p的右子树作为r
的右子树。
2。若p没有左子树，直接用p的右孩子取代它。
两种方法各有优劣，第一种操作简单一点点，但均衡性不如第二种，因为它将结点p的右子树
全部移到左边来了。下面将分别以两种种思路编写代码。

第一种：
btree * DelNode(btree *p)
{
if (p->lchild)
{
btree *r = p->lchild; //r指向其左子树;
while(r->rchild != NULL)//搜索左子树的最右边的叶子结点r
{
r = r->rchild;
}
r->rchild = p->rchild;

btree *q = p->lchild; //q指向其左子树;
free(p);
return q;
}
else
{
btree *q = p->rchild; //q指向其右子树;
free(p);
return q;
}
}

第二种：
btree * DelNode(btree *p)
{
if (p->lchild)
{
btree *r = p->lchild; //r指向其左子树;
btree *prer = p->lchild; //prer指向其左子树;
while(r->rchild != NULL)//搜索左子树的最右边的叶子结点r
{
prer = r;
r = r->rchild;
}

if(prer != r)//若r不是p的左孩子，把r的左孩子作为r的父亲的右孩子
{
prer->rchild = r->lchild;
r->lchild = p->lchild; //被删结点p的左子树作为r的左子树
}
r->rchild = p->rchild; //被删结点p的右子树作为r的右子树

free(p);
return r;
}
else
{
btree *q = p->rchild; //q指向其右子树;
free(p);
return q;
}
}
但是上面这种方法，把r移来移去，很容易出错，其实在这里我们删除的只是p的元素值，而不是它的地址，所以完全没有必要移动指针。仔细观察，发现我们删除的地址实际上是p的左子树的最右边的叶子结点r的地址，所以我们只要把r的数据填到p中，然后把r删除即可。
算法如下：
btree * DelNode(btree *p)
{
if (p->lchild)
{
btree *r = p->lchild; //r指向其左子树;
btree *prer = p->lchild; //prer指向其左子树;
while(r->rchild != NULL)//搜索左子树的最右边的叶子结点r
{
prer = r;
r = r->rchild;
}
p->data = r->data;

if(prer != r)//若r不是p的左孩子，把r的左孩子作为r的父亲的右孩子
prer->rchild = r->lchild;
else
p->lchild = r->lchild; //否则结点p的左子树指向r的左子树

free(r);
return p;
}
else
{
btree *q = p->rchild; //q指向其右子树;
free(p);
return q;
}
}

转http://www.programfan.com/

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greaper615 2012-10-15

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嗯学习了。原来只要直接交换值就可以了，这都没想到，我真是笨死了……

qirutr 2012-10-12

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学习，学习的，顶顶！！！！

牙牙学语 2012-10-10

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mark

xingpeiyue 2012-10-09

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喜欢，很想学

magic_an001 2012-10-08

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mark，学习

花花花湖 2012-10-08

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学习了。不过，数据结构上的那个方法挺好的。

卡卡_苏米 2011-11-23

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很好，高手~·

yinpan1990 2011-11-22

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bomiaomiao123 2011-11-19

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mark

小步舞曲 2011-09-17

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[Quote=引用 17 楼 zfzf294990051 的回复:]
typedef int Status;
Status Delete(BiTree &T)
{
if(!p->rchild)//右子树空，只需重接左子树
{
q=p;p=p->lchild;free(q);
}
else if(!p->lchild)//左子树空，只需接右子树
q=p;p=p->rchild;free(q);
else//左右子树均不空
{
q=p;
……
[/Quote]

q=p;p=p->lchild;free(q);
为什么只需重新接子数？如果这样那p的父节点的lChild或者rChild的值会变成什么？？
这样不是也变成森林了嘛？？

INGDI 2010-05-24

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mark
顶27楼

fairywell 2010-05-22

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恩，写的不错，可以参考下。
目前数据结构里常用的方法应该是，对待删除节点有两个孩子的情况，用该节点的右子树的最小元素取代这个节点，然后删除这个右子树里的最小元素即可。
pseudocode如下：



if (T->Left && T-Right) /* two children */

{

    TmpCell = FindMin(T->Right);

    T->Element = TmpCell->Element;

    T->Right = Delete(T->Element, T->Right);

}

STL 源代码里面我记得也是用的这个方法

hellodota121 2010-05-22

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这个东西肯定要MARK
为了以后的提升

azsx18726 2010-05-22

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学习！

Sunday 2010-05-19

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http://zhuyingqingfen.download.csdn.net/

yaoweijq 2010-03-19

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java版的一种方案，另一种没写代码实现过。。。



/**

	 * 1。若p有左子树，用p的左孩子取代它；

	 * 找到其左子树的最右边的叶子结点r，

	 * 把p的右子树作为r的右子树。

     * 2。若p没有左子树，直接用p的右孩子取代它。

	 * @param key

	 */

	public void deleteValue(T key){

		if(key == null){

			return;

		}

		BinTreeNode<T>node = new BinTreeNode<T>(key);

		if(this.getRootNode() == null || this.getRootNode().getValue() == null){

			return;

		}

		BinTreeNode<T>parentNode = this.getRootNode();

		BinTreeNode<T>tempNode = this.getRootNode();

		while(true){

			if(tempNode.compareTo(node) == 0){

				//找到要删除的结点了

				//case 1 tempNode有左子树

				if(tempNode.getLeftNode()!=null){

					BinTreeNode<T>leftNode = tempNode.getLeftNode();

					BinTreeNode<T>rightNode = tempNode.getRightNode();

					BinTreeNode<T>subLeftNode = leftNode.getLeftNode();

					BinTreeNode<T>subRightNode = leftNode.getRightNode();

					tempNode.setValue(leftNode.getValue());

					leftNode.setLeftNode(null);

					leftNode.setRightNode(null);

					leftNode = null;

					tempNode.setLeftNode(subLeftNode);

					tempNode.setRightNode(subRightNode);

					while(tempNode.getRightNode()!=null){

						tempNode = tempNode.getRightNode();

					}

					tempNode.setRightNode(rightNode);

				}

				//case 2 tempNode无左子树

				else{

					if(tempNode.getRightNode()==null){

						if(parentNode.compareTo(this.getRootNode())==0){

							this.rootNode = null;

							return;

						}

						if(parentNode.getLeftNode()!=null&&parentNode.getLeftNode().compareTo(tempNode)==0){

							parentNode.setLeftNode(null);

						}else{

							parentNode.setRightNode(null);

						}

						tempNode = null;

					}else{

						BinTreeNode<T>leftNode = tempNode.getRightNode().getLeftNode();

						BinTreeNode<T>rightNode = tempNode.getRightNode().getRightNode();

						BinTreeNode<T>tempRightNode = tempNode.getRightNode();

						tempNode.setLeftNode(leftNode);

						tempNode.setRightNode(rightNode);

						tempNode.setValue(tempRightNode.getValue());

						tempRightNode.setLeftNode(null);

						tempRightNode.setRightNode(null);

						tempRightNode = null;

					}

				}

				//删除end

				return;

			}else if(tempNode.compareTo(node)>0){

				if(tempNode.getLeftNode()==null){

					return;

				}else{

					parentNode = tempNode;

					tempNode = tempNode.getLeftNode();

				}

			}else{

				if(tempNode.getRightNode()==null){

					return;

				}else{					

					parentNode = tempNode;

					tempNode = tempNode.getRightNode();

				}

			}

		}

	}