c++二叉查找树删除

Question

所以，我的问题是我不明白为什么这不起作用。我在下面评论说，明明是初始化的，却从来没有初始化。我是不是把指针写错了，我是不是把逻辑搞反了我是不是偏离太远了，还是从头开始更好?这是我遇到的最难的任务，所以任何帮助都将是非常有益的。

void Dictionary::remove(string word)
{
if(root == NULL)
{
    cout << "list is emptyn";
    return;
}
DictionaryNode *curr = root;
DictionaryNode *parent = NULL;`
while(curr != NULL)
{
    if(curr->word == word)
        break;
    else
    {
        parent = curr;
        if(word > curr->word)
            curr = curr->right;
        else
            curr = curr->left;
    }
}
//LEAF node.
if(curr->left == NULL && curr->right == NULL)
{
    if(parent->left == curr) // Right here is an access violation. Which doesn't //make sense.
    {
        parent->left = NULL;
    }
    else
    {
        parent->right = NULL;
    }
    delete curr;
}
/*
* Node has a single child LEFT or RIGHT
*/  
if((curr->left == NULL && curr->right != NULL) || (curr->left != NULL && curr->right == NULL))
{
    if(curr->left == NULL && curr->right != NULL)
    {
        if(parent->left == curr) //if(parent->left == curr) //says parent is //not intialized
        {
            parent->left = curr->right;
            delete curr;
        }
        else
        {
            parent->right = curr->right;
            delete curr;
        }
    }
    else
    {
        if(parent->left == curr)
        {
            parent->left = curr->left;
            delete curr;
        }
        else
        {
            parent->right = curr->left;
            delete curr;
        }
    }
}
 if (curr->left != NULL && curr->right != NULL)
{
    DictionaryNode* temp; 
    if(parent == NULL || parent->left==curr)
    {  
        temp = curr->right;
        while(temp->left!=NULL)
            temp = temp->left;
        if(parent!=NULL)
            parent->left = curr->right;
        else
            root = curr->right;
        temp->left = curr->left;
        curr->left = curr->right=NULL;
        delete curr;
    } 
    else if(parent->right==curr)
    {
        temp = curr->left;
        while(temp->right!=NULL)
            temp = temp->right;
        parent->right=curr->left;
        temp->right = curr->right;
        curr->left = curr->right=NULL;
        delete curr;
    }
  }
}

Answer 1

1。我看到的第一件事:

while(curr != NULL)
{
 //stuff
} 

正如所写的，似乎在你的循环结束时curr == NULL

Lazy me必须查看循环的内容才能注意到中断。如果循环中有一个更大的块，那么中断就更不明显了。这不是一个好的做法。

使用bool(例如:bool isNodeFound;)，它便宜(一个位)并且使它更清晰。= NULL &&!isNodeFound)乍一看更清楚您的意图，而无需查看循环的内容。

2。如果您确实没有在循环中击中断点并且curr == NULL怎么办?您的下一个指令(左)将失败!看来布尔值又会派上用场了!

if(!isNodeFound)
{
//log error if you can "cannot remove node because it is not in dictionary"
return false; //make your function a bool to return if it succeeded or not
}

试着用同样的心态分析你的代码的其余部分，更清晰和测试，让我知道它是否有效。

Answer 2

大家。有一天，当我需要在BST中删除树节点的函数时，我搜索了这个问题。所以，这个问题很好，我编辑并检查了上面的代码，然后代码真的运行成功。上面的代码遗漏了一些实例，请跟随我下面的解释:

首先，删除的节点是LEAF node。你错过了一个节点是根节点或叶节点的实例(即BST只有一个节点)。因此，parent为NULL, parent->left/right无效。

第二，被删除的节点在左侧或右侧有一个子树。因此，如果删除的节点是根节点，这与First类似。

第三，删除节点有左、右子树。您考虑了"parent"，但您不应该使用"if(parent == NULL || parent->left==curr)"，就好像parent = NULL一样，以便parent->left无效。你应该让"if(parent == NULL){…} else{if(parent->left == curr)…}".

最后，使用if…else-if…Else代替if…if…如果因为你删除了"curr"，那么你将不知道"curr"指向任何地方，" If "接下来仍然会被检查为"curr"错误。

下面编辑的代码为任何人需要，

void Dictionary::remove(string word)
{
    if(root == NULL)
    {
        cout << "list is emptyn";
        return;
    }
    DictionaryNode *curr = root;
    DictionaryNode *parent = NULL;
    while(curr != NULL)
    {
        if(curr->word == word)
            break;
        else
        {
            parent = curr;
            if(word > curr->word)
                curr = curr->right;
            else
                curr = curr->left;
        }
    }
    //LEAF node.
    if(curr->left == NULL && curr->right == NULL)
    {
        if (parent == NULL) {
            delete curr;
        } else {
            if(parent->left == curr) // Right here is an access violation. Which doesn't //make sense.
            {
                parent->left = NULL;
            }
            else
            {
                parent->right = NULL;
            }
            delete curr;
        }
    }
    /*
    * Node has a single child LEFT or RIGHT
    */  
    else if((curr->left == NULL && curr->right != NULL) || (curr->left != NULL && curr->right == NULL))
    {
        if(curr->left == NULL && curr->right != NULL)
        {
            if (parent == NULL) {
                    root = curr->right;
                    curr->right = NULL;
                    delete curr;
            } else {
                if(parent->left == curr) //if(parent->left == curr) //says parent is //not intialized
                {
                    parent->left = curr->right;
                    delete curr;
                }
                else
                {
                    parent->right = curr->right;
                    delete curr;
                }
            }
        }
        else
        {
            if (parent == NULL) {
                    root = curr->left;
                    curr->left = NULL;
                    delete curr;
            } else {
                if(parent->left == curr)
                {
                    parent->left = curr->left;
                    delete curr;
                }
                else
                {
                    parent->right = curr->left;
                    delete curr;
                }
            }
        }
    }
    else
    {
        DictionaryNode* temp; 
        if(parent == NULL)
        {  
            temp = curr->right;
            while(temp->left!=NULL)
                temp = temp->left;
            if(parent!=NULL)
                parent->left = curr->right;
            else
                root = curr->right;
            temp->left = curr->left;
            curr->left = curr->right=NULL;
            delete curr;
        } else {
            if(parent->left==curr){
                temp = curr->right;
                while(temp->left!=NULL)
                    temp = temp->left;
                if(parent!=NULL)
                    parent->left = curr->right;
                else
                    root = curr->right;
                temp->left = curr->left;
                curr->left = curr->right=NULL;
                delete curr;
            }
            else if(parent->right==curr)
            {
                temp = curr->left;
                while(temp->right!=NULL)
                    temp = temp->right;
                parent->right=curr->left;
                temp->right = curr->right;
                curr->left = curr->right=NULL;
                delete curr;
            }
        }
    }
}

希望这段代码能在别人需要的时候帮助到他们!