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一步一步写算法(之排序二叉树删除-3)(一)
2014-11-23 23:36:33 来源: 作者: 【 】 浏览:10
Tags:步一步 算法 排序 删除

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3 普通节点的删除

3.1 删除的节点没有左子树,也没有右子树

测试用例1: 删除节点6

/*

*

* 10 ======> 10

* / \ \

* 6 15 15

*

*/

static void test8()

{

TREE_NODE* pTreeNode = NULL;

assert(TRUE == insert_node_into_tree(&pTreeNode, 10));

assert(TRUE == insert_node_into_tree(&pTreeNode, 6));

assert(6 == pTreeNode->left_child->data);

assert(TRUE == insert_node_into_tree(&pTreeNode, 15));

assert(TRUE == delete_node_from_tree(&pTreeNode, 6));

assert(NULL == pTreeNode->left_child);

free(pTreeNode->right_child);

free(pTreeNode);

}

/*

*

* 10 ======> 10

* / \ \

* 6 15 15

*

*/

static void test8()

{

TREE_NODE* pTreeNode = NULL;

assert(TRUE == insert_node_into_tree(&pTreeNode, 10));

assert(TRUE == insert_node_into_tree(&pTreeNode, 6));

assert(6 == pTreeNode->left_child->data);

assert(TRUE == insert_node_into_tree(&pTreeNode, 15));

assert(TRUE == delete_node_from_tree(&pTreeNode, 6));

assert(NULL == pTreeNode->left_child);

free(pTreeNode->right_child);

free(pTreeNode);

} 测试用例2: 删除节点15

/*

*

* 10 ======> 10

* / \ /

* 6 15 6

*

*/

static void test9()

{

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, 15));

assert(15 == pTreeNode->right_child->data);

assert(TRUE == delete_node_from_tree(&pTreeNode, 15));

assert(NULL == pTreeNode->right_child);

free(pTreeNode->right_child);

free(pTreeNode);

}

/*

*

* 10 ======> 10

* / \ /

* 6 15 6

*

*/

static void test9()

{

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, 15));

assert(15 == pTreeNode->right_child->data);

assert(TRUE == delete_node_from_tree(&pTreeNode, 15));

assert(NULL == pTreeNode->right_child);

free(pTreeNode->right_child);

free(pTreeNode);

} 那么代码应该怎么编写呢?

STATUS _delete_node_from_tree(TREE_NODE* pTreeNode)

{

TREE_NODE* pLeftMax;

if(NULL == pTreeNode-> left_child && NULL == pTreeNode->right_child){

if(pTreeNode == pTreeNode->parent->left_child)

pTreeNode->parent->left_child = NULL;

else

pTreeNode->parent->right_child = NULL;

}

free(pTreeNode);

return TRUE;

}

STATUS _delete_node_from_tree(TRE

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