函数从二叉搜索树中删除具有给定值的节点

Question

我有一个名为TreeNode的结构，它有一个int键、左键、右键和父键。我正试图用DeleteNode函数从树中删除一个节点，但它不起作用。我应该用左边子树中的最大值替换DeleteNode函数中已删除的节点。我的移植和max函数是deletenode的助手函数。我的问题是，我不确定在DeleteNode函数中，我应该将我所在的节点值与通过函数传入的值进行比较。我在代码中有一个注释，我不知道该怎么做。任何帮助都将不胜感激！

void transplant(TreeNode* u, TreeNode* v)   //swaps u with v
{

if (u->parent == NULL)  //if u was root, make v new root
    u->parent = v;
else if (u == u->parent->left)  //if u is smaller than it's parent
    u->parent->left = v;        //set v to the left child of parent of u. Swap them at left, really
else
    u->parent->right = v;       //otherwise swap them at right

if (v != NULL)              //reassign parents to double link
    v->parent = u->parent;
}

TreeNode* maximum(TreeNode* n)
{

while (n->left != NULL)
    n = n->left;
return n;
}

  void deleteNode(TreeNode *node, int key)
{

if (node->left == NULL)                 //if there is no left child
    transplant(node, node->right);      //swap 
else if (node->right == NULL)           //if there is no right child
    transplant(node, node->left);       //swap 
else 
{
  if(node->key == key){ //****This if comparison must be wrong***
    TreeNode* temp = maximum(node->right);  //make temp the max on right
    if (temp->parent != node )              //if it is more than one chain down
    {
        transplant(temp, temp->right);          //swap temp and it's right branch
        temp->right = node->right;          //set right branch to nodes right
        temp->parent->right = temp;             //set temp to the right child 
    }
    transplant(node, temp);                 // transplant
    temp->left = node->left;                //get nodes left branch
    temp->left->parent = temp;              //replace
    }
}
}

Answer 1

首先，您有三个案例需要处理：（直接从维基百科。当我学习数据结构时，这对我来说很好）

有三种可能的情况需要考虑：

删除一个没有子节点（叶）：只需从树中删除该节点。

由于最大值（node->right)，您似乎试图实现按顺序后继选项。

现在我们已经确定了你可能的病例，唯一真正需要移植的病例是第三例，在我看来有两个孩子。

情况1：只需在叶节点上调用delete即可。

del->right->parent = del->parent;
if (del == del->parent->left)
    del->parent->left = del->right;
else if (del == del->parent->right)
    del->parent->right = del->right;
delete del;

TreeNode *inOrderSuccessor = maximum(del->right);
del->val = inOrderSuccessor->val; //you could use transplant/swap here
deleteNode(inOrderSuccessor, inOrderSuccessor->val);