Question

所以这是我到目前为止的代码。我认为我大部分时间都有逻辑，但我有两个孩子的问题。我已经弄清楚我必须使用什么样的算法才能使它工作（我用注释注释）在功能的底部），我只是不确定如何开始......有人能指出我正确的方向吗？

到目前为止我所拥有的：

void BinaryTree::remove(int data){
// Is the data valid?
if(data < 0){
    cerr << data << " is not valid.  Must be a positive integer value."  << endl;
}
else{
    // Find the node
    BinNode* loc = root;
    BinNode* parent = nullptr;
    bool found = false;
    while(!found && loc != nullptr){
        if(data > loc->data){
            parent = loc;
            loc = loc->right;
        }
        else if(data < loc->data){
            parent = loc;
            loc = loc->left;
        }
        else found = true;
    }

    // If there is a parent, take care of the pointer
    if(parent != nullptr){
        if(loc->data < parent->data)
            parent->left = nullptr;
        else if(loc->data > parent->data)
            parent->right = nullptr;
    }
    // If there are children, save pointers to them
    BinNode* leftChild = nullptr;
    BinNode* rightChild = nullptr;  
    if(loc->left != nullptr)
        leftChild = loc->left;
    if(loc->right != nullptr)
        rightChild = loc->right;

    // So now pointers to the children have been saved (if they exist) and
    // parent pointers have been taken care of (if they exist) the node can be deleted
    // If no children exist simply just delete the node and return
    delete loc;
    // If one child exists
    if(leftChild != nullptr || rightChild != nullptr){  
        if(leftChild != nullptr){
            if(leftChild->data < parent->data)
                parent->left = leftChild;
            else if(leftChild->data > parent->data)
                parent->right = leftChild;
        }
        else if(rightChild != nullptr){
            if(rightChild->data < parent->data)
                parent->left = rightChild;
            else if(rightChild->data > parent->data)
                parent->right = rightChild;
        }       
    }

    // Both children exist...this sucks.
    else if(leftChild != nullptr && rightChild != nullptr){
        // Find a minimum in the right subtree
        BinNode * min = root;
        BinNode * minParent = nullptr;
        while(min->left != nullptr){
            minParent = min;
            min = min->left;
        }
        // Replace value of the node to be removed with the found minimum
        loc = new BinNode(min->data);

        // Delete the remaining duplicate node
        if(minParent != nullptr)
            minParent->left = nullptr;
        delete min;
    }
}
}

Answer 1

我们在这里。这似乎有效。如果有人对我如何优化这个或我需要修复的问题有任何建议请说出来！谢谢！

void BinaryTree::remove(int data){
// Is the data valid?
if(data < 0){
    cerr << data << " is not valid.  Must be a positive integer value."  << endl;
}
else{
    // Find the node
    BinNode* loc = root;
    BinNode* parent = nullptr;
    bool found = false;
    while(!found && loc != nullptr){
        if(data > loc->data){
            parent = loc;
            loc = loc->right;
        }
        else if(data < loc->data){
            parent = loc;
            loc = loc->left;
        }
        else found = true;

    // If there are children, save pointers to them
    BinNode* leftChild = nullptr;
    BinNode* rightChild = nullptr;  
    if(loc->left != nullptr)
        leftChild = loc->left;
    if(loc->right != nullptr)
        rightChild = loc->right;

    // So now pointers to the children have been saved (if they exist) and
    // parent pointers have been taken care of (if they exist) the node can be deleted
    // If no children exist simply just delete the node and return

    // Check if two children exist
    if(leftChild != nullptr && rightChild != nullptr){

        // Find a minimum in the right subtree
        BinNode * min = loc->right;
        BinNode * minParent = loc;
        while(min->left != nullptr){
            minParent = min;
            min = min->left;
        }
        // Replace value of the node to be removed with the found minimum
        loc->data = min->data;

        // Delete the duplicate
        if(minParent != loc)
            minParent->left = nullptr;
        else minParent->right = nullptr;
        delete min;
    }

    // If one child exists
    // Need to handle if it is the root here
    // change root pointer to remaining child
    else if(leftChild != nullptr || rightChild != nullptr){ 
        // If there is a parent, take care of the pointer
        if(parent != nullptr){
            if(loc->data < parent->data)
                parent->left = nullptr;
            else if(loc->data > parent->data)
                parent->right = nullptr;

        // Now loc can be deleted
        delete loc;
        if(leftChild != nullptr){
            if(parent != nullptr){
                if(leftChild->data < parent->data)
                    parent->left = leftChild;
                else if(leftChild->data > parent->data)
                    parent->right = leftChild;
            }
            // If it is the root
            else{
                root = leftChild;
            }
        }
        else if(rightChild != nullptr){
            if(parent != nullptr){
                if(rightChild->data < parent->data)
                    parent->left = rightChild;
                else if(rightChild->data > parent->data)
                parent->right = rightChild;
            }
            // If it is the root
            else{
                root = rightChild;
            }
        }       
    }

    // If there are no children
    else if (leftChild == nullptr && rightChild == nullptr){
        // If there is a parent, take care of the pointer
        if(parent != nullptr){
            if(loc->data < parent->data)
                parent->left = nullptr;
            else if(loc->data > parent->data)
                parent->right = nullptr;
        }
        // If it is the root
        else 
            root = nullptr;
        delete loc;
    }
}

删除二叉树中具有两个子节点的节点

1 个答案: