爲什麼在NULL檢查之後會在指針上調用遞歸函數？

Question

// So I call this function after deleting a node. 
// It works before I delete the node, but for some reason 
// after I perform a deletion and update the tree it runs into 
// EXC_BAD_ACCESS on the line below... 

void BinaryTree::updateCost(BinaryNode *root) { 
    if (root != NULL) 
     root->updateCostRecursively(1); 
} 

void BinaryNode::updateCostRecursively(int newCost) { 
    cout << this << endl; // prints 0x3000000000000000 before the bad access 
    cost = newCost; // has a bad access here 
    if (right != NULL) 
     right->updateCostRecursively(newCost + 1); 
    if (left != NULL) 
     left->updateCostRecursively(newCost + 1); 
} 

爲什麼在NULL對象上調用這個遞歸函數，即使每次都檢查指針？

我已經複製了我用來刪除下面節點的代碼。我仍然無法理解遞歸函數，但是從什麼時候我都可以告訴我離開一個懸掛的指針。

BinaryNode *BinaryTree::findMin(BinaryNode *t) { 
    if (t == NULL) return NULL; 
    while (t->left != NULL) t = t->left; 
    return t; 
} 

BinaryNode *BinaryTree::removeMin(BinaryNode *t) { 
    if (t == NULL) return NULL; 
    if (t->left != NULL) 
     t->left = removeMin(t->left); 
    else { 
     BinaryNode *node = t; 
     t = t->right; 
     delete node; 
    } 
    return t; 
} 

bool BinaryTree::remove(int key) { 
    if (root != NULL && remove(key, root)) 
     return true; 
    return false; 
} 

BinaryNode *BinaryTree::remove(int x, BinaryNode *t) { 
    if (t == NULL) return NULL; 

    if (x < t->key) 
     t->left = remove(x, t->left); 
    else if (x > t->key) 
     t->right = remove(x, t->right); 
    else if (t->left != NULL && t->right != NULL) { 
     // item x is found; t has two children 
     t->key = findMin(t->right)->key; 
     t->right = removeMin(t->right); 
    } else { //t has only one child 
     BinaryNode *node = t;  
     t = (t->left != NULL) ? t->left : t->right; 
     delete node; 
    } 

    updateCost(root); 
    return t; 
} 

Answer 1

錯誤發生在您的刪除方法中，而不是您發佈的代碼。刪除節點（例如root->right）後，需要設置root->right = NULL。你正在用delete做的事情是釋放指針指向的內存。指針本身繼續指向該地址。您正在嘗試訪問已釋放的內存，因此出現訪問異常錯誤。