排序二叉树删除-2

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