一步一步写算法（之排序二叉树删除-2）

原文:一步一步写算法（之排序二叉树删除-2）

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    2.4 删除节点的左右子树都存在，此时又会分成两种情形

    1）左节点是当前左子树的最大节点，此时只需要用左节点代替根节点即可

/*
*
*         10          ======>     6
*        /  \                   /   \
*      6     15               5     15
*     /
*    5
*/

    代码该怎么编写呢？

STATUS delete_node_from_tree(TREE_NODE** ppTreeNode, int data)
{
	TREE_NODE* pTreeNode;
	TREE_NODE* pLeftMax;

	if(NULL == ppTreeNode || NULL == *ppTreeNode)
		return FALSE;

	pTreeNode = find_data_in_tree_node(*ppTreeNode, data);
	if(NULL == pTreeNode)
		return FALSE;

	if(*ppTreeNode == pTreeNode){

		if(NULL == pTreeNode->left_child && NULL == pTreeNode->right_child){
			*ppTreeNode = NULL;
		}else if(NULL != pTreeNode->left_child && NULL == pTreeNode->right_child){
			*ppTreeNode = pTreeNode->left_child;
			pTreeNode->left_child->parent = NULL;
		}else if(NULL == pTreeNode->left_child && NULL != pTreeNode->right_child){
			*ppTreeNode = pTreeNode->right_child;
			pTreeNode->right_child->parent = NULL;
		}else{
			pLeftMax = find_max_node(pTreeNode->left_child);
			if(pLeftMax == pTreeNode->left_child){
				*ppTreeNode = pTreeNode->left_child;
				(*ppTreeNode)->right_child = pTreeNode->right_child;
				(*ppTreeNode)->right_child->parent = *ppTreeNode;
				(*ppTreeNode)->parent = NULL;
			}
		}

		free(pTreeNode);
		return TRUE;
	}

	return TRUE;
}

    上面的代码中添加的内容表示了我们介绍的这一情形。为此，我们可以设计一种测试用例。依次插入10、6、5、15，然后删除10即可。

static void test6()
{
	TREE_NODE* pTreeNode = NULL;
	assert(TRUE == insert_node_into_tree(&pTreeNode, 10));
	assert(TRUE == insert_node_into_tree(&pTreeNode, 6));
	assert(TRUE == insert_node_into_tree(&pTreeNode, 5));
	assert(TRUE == insert_node_into_tree(&pTreeNode, 15));
	assert(TRUE == delete_node_from_tree(&pTreeNode, 10));
	assert(6 == pTreeNode->data);
	assert(NULL == pTreeNode->parent);
	assert(15 == pTreeNode->right_child->data);
	assert(pTreeNode = pTreeNode->right_child->parent);
	assert(NULL == pTreeNode->parent);
	free(pTreeNode->left_child);
	free(pTreeNode->right_child);
	free(pTreeNode);
}

    如果上面的测试用例通过，表示我们添加的代码没有问题。

    2）左节点不是当前左子树的最大节点，情形如下所示

/*
*
*         10          ======>     8
*        /  \                   /   \
*      6     15               5     15
*       \
*        8
*/

    此时，我们应该用10左侧的最大节点8代替删除的节点10即可。

STATUS delete_node_from_tree(TREE_NODE** ppTreeNode, int data)
{
	TREE_NODE* pTreeNode;
	TREE_NODE* pLeftMax;

	if(NULL == ppTreeNode || NULL == *ppTreeNode)
		return FALSE;

	pTreeNode = find_data_in_tree_node(*ppTreeNode, data);
	if(NULL == pTreeNode)
		return FALSE;

	if(*ppTreeNode == pTreeNode){

		if(NULL == pTreeNode->left_child && NULL == pTreeNode->right_child){
			*ppTreeNode = NULL;
		}else if(NULL != pTreeNode->left_child && NULL == pTreeNode->right_child){
			*ppTreeNode = pTreeNode->left_child;
			pTreeNode->left_child->parent = NULL;
		}else if(NULL == pTreeNode->left_child && NULL != pTreeNode->right_child){
			*ppTreeNode = pTreeNode->right_child;
			pTreeNode->right_child->parent = NULL;
		}else{
			pLeftMax = find_max_node(pTreeNode->left_child);
			if(pLeftMax == pTreeNode->left_child){
				*ppTreeNode = pTreeNode->left_child;
				(*ppTreeNode)->right_child = pTreeNode->right_child;
				(*ppTreeNode)->right_child->parent = *ppTreeNode;
				(*ppTreeNode)->parent = NULL;
			}else{
				pTreeNode->data = pLeftMax->data;
				pLeftMax->parent->right_child = NULL;
				pTreeNode = pLeftMax;
			}
		}

		free(pTreeNode);
		return TRUE;
	}

	return TRUE;
}

    那么，这个场景下面测试用例又该怎么设计呢？其实只需要按照上面给出的示意图进行即可。依次插入数据10、6、8、15，然后删除数据10。

static void test7()
{
	TREE_NODE* pTreeNode = NULL;
	assert(TRUE == insert_node_into_tree(&pTreeNode, 10));
	assert(TRUE == insert_node_into_tree(&pTreeNode, 6));
	assert(TRUE == insert_node_into_tree(&pTreeNode, 8));
	assert(TRUE == insert_node_into_tree(&pTreeNode, 15));
	assert(TRUE == delete_node_from_tree(&pTreeNode, 10));
	assert(8 == pTreeNode->data);
	assert(NULL == pTreeNode->parent);
	assert(NULL == pTreeNode->left_child->right_child);
	assert(NULL == pTreeNode->parent);
	free(pTreeNode->left_child);
	free(pTreeNode->right_child);
	free(pTreeNode);
}

    至此，删除节点为根节点的情形全部讨论完毕，那么如果删除的节点是普通节点呢，那应该怎么解决呢？

STATUS delete_node_from_tree(TREE_NODE** ppTreeNode, int data)
{
	TREE_NODE* pTreeNode;
	TREE_NODE* pLeftMax;

	if(NULL == ppTreeNode || NULL == *ppTreeNode)
		return FALSE;

	pTreeNode = find_data_in_tree_node(*ppTreeNode, data);
	if(NULL == pTreeNode)
		return FALSE;

	if(*ppTreeNode == pTreeNode){

		if(NULL == pTreeNode->left_child && NULL == pTreeNode->right_child){
			*ppTreeNode = NULL;
		}else if(NULL != pTreeNode->left_child && NULL == pTreeNode->right_child){
			*ppTreeNode = pTreeNode->left_child;
			pTreeNode->left_child->parent = NULL;
		}else if(NULL == pTreeNode->left_child && NULL != pTreeNode->right_child){
			*ppTreeNode = pTreeNode->right_child;
			pTreeNode->right_child->parent = NULL;
		}else{
			pLeftMax = find_max_node(pTreeNode->left_child);
			if(pLeftMax == pTreeNode->left_child){
				*ppTreeNode = pTreeNode->left_child;
				(*ppTreeNode)->right_child = pTreeNode->right_child;
				(*ppTreeNode)->right_child->parent = *ppTreeNode;
				(*ppTreeNode)->parent = NULL;
			}else{
				pTreeNode->data = pLeftMax->data;
				pLeftMax->parent->right_child = pLeftMax->left_child;
				pLeftMax->left_child->parent = pLeftMax->parent;
				pTreeNode = pLeftMax;
			}
		}

		free(pTreeNode);
		return TRUE;
	}

	return _delete_node_from_tree(pTreeNode);
}

    我们在当前函数的最后一行添加_delete_node_from_tree，这个函数用来处理普通节点的删除情况，我们会在下面一篇博客中继续介绍。

    3、 普通节点的删除

    （待续）