https://www.geeksforgeeks.org/splay-tree-set-1-insert/

When I was learning about splay trees, I referred to this article and had a little doubt about the code in it.

this is splay function in this article

struct node *splay(struct node *root, int key)
{
    // Base cases: root is NULL or key is present at root
    if (root == NULL || root->key == key)
        return root;
 
    // Key lies in left subtree
    if (root->key > key)
    {
        // Key is not in tree, we are done
        if (root->left == NULL) return root;
 
        // Zig-Zig (Left Left)
        if (root->left->key > key)
        {
            // First recursively bring the key as root of left-left
            root->left->left = splay(root->left->left, key);
 
            // Do first rotation for root, second rotation is done after else
            root = rightRotate(root);
        }
        else if (root->left->key < key) // Zig-Zag (Left Right)
        {
            // First recursively bring the key as root of left-right
            root->left->right = splay(root->left->right, key);
 
            // Do first rotation for root->left
            if (root->left->right != NULL)
                root->left = leftRotate(root->left);
        }
 
        // Do second rotation for root
        return (root->left == NULL)? root: rightRotate(root);
    }
    else // Key lies in right subtree
    {
        // Key is not in tree, we are done
        if (root->right == NULL) return root;
 
        // Zag-Zig (Right Left)
        if (root->right->key > key)
        {
            // Bring the key as root of right-left
            root->right->left = splay(root->right->left, key);
 
            // Do first rotation for root->right
            if (root->right->left != NULL)
                root->right = rightRotate(root->right);
        }
        else if (root->right->key < key)// Zag-Zag (Right Right)
        {
            // Bring the key as root of right-right and do first rotation
            root->right->right = splay(root->right->right, key);
            root = leftRotate(root);
        }
 
        // Do second rotation for root
        return (root->right == NULL)? root: leftRotate(root);
    }
}

Here is a snippet from the source code return (root->left == NULL)? root: rightRotate(root);, it appears in the last paragraph of the function. My question is if I can write like thisreturn rightRotate(root), Because we have already judged whether this pointer is NULL above, It seems to me that after we do the rotate operation, the root->left can not be NULL. If this pointer can be null, hope to point out the situation that does not hold

CodePudding user response：

is this Ternary expression redundant in splay function

No, root = rightRotate(root); and root = leftRotate(root); change root.