太惡心了，三種情況理解就好

//! 刪除結點 void BinarySearchTreesZH::remove(Node *node) {if (node->left != nullptr && node->right != nullptr)//! 把刪除度為2 的結點的問題轉化為刪除{node->element = predecessor(node)->element;remove(predecessor(node));}else if (node->left == nullptr && node->right == nullptr) //! 刪除度為0的結點{if (node->parent == nullptr) //! 若結點是根結點直接置空{node = nullptr;}if (node == node->parent->left) //!結點時葉結點時的置空{node->parent->left = nullptr;}if (node == node->parent->right){node->parent->right = nullptr;}//!我覺得這里直接全都node = nullptr應該是一樣的}else{//! 刪除度為1的結點if (node->parent == nullptr)//!若結點是根結點，用其子結點替代成為根結點{if (node->left != nullptr){root = node->left;}else if (node->right != nullptr){root = node->right;}}else if (node == node->parent->left){//! 結點是父結點的左子結點或右子結點時，改動雙向指針，和雙向鏈表刪除結點操作一樣if (node->left != nullptr){node ->parent->left = node ->left;node ->left->parent = node ->parent;}else if (node->right != nullptr){node ->parent->left = node->left;node ->left ->parent = node ->parent;}}else if (node == node ->parent ->right){if (node->left != nullptr){node ->parent->right = node ->left;node ->left->parent = node ->parent;}else if (node->right != nullptr){node ->parent->right = node->left;node ->left ->parent = node ->parent;}}} }

總結

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