std/container/rbtree.d

120      * Get the right child
122     @property inout(RBNode)* right() inout return scope
153      * Set the right child.  Also updates the new child's parent node.  This
161     @property Node right(return scope Node newNode) @trusted
171     // performs rotate-right operation, where this is T, _right is R, _left is
183      * Rotate right.  This performs the following operations:
185      *  - This node becomes the new parent's right child.
186      *  - The old right child of the new parent becomes the left child of this
200             parent.right = _left;
204         _left.right = &this;
227      *  - The right child becomes the parent of this node.
229      *  - The old left child of the new parent becomes the right child of this
235         assert(_right !is null, "right node must not be null");
243             parent.right = _right;
250         right = tmp;
327                         // parent is right node, y is 'uncle'
412                 _parent.right = y;
418                 y.right = &this;
420                 y.right = _right;
427             right = yr;
433                     yp.right = &this;
454             _parent.right = x;
474                     Node wr = w.right;
501                     // right node
511                     Node wr = w.right;
548                 _parent.right = null;
589         if (n.right is null)
596             return n.right.leftmost;
631             copy.right = _right.dup(alloc);
642             copy.right = _right.dup();
857                     cur = cur.right;
875                     cur = cur.right;
913                         // add to right of new parent
931                     nxt = newParent.right;
935                         // add to right of new parent
938                         (() @trusted { newParent.right = result; }) ();
1642                 cur = cur.right;
1656                 cur = cur.right;
1772                 printTree(n.right, indent + 2);
1802                 if (n.parent.left !is n && n.parent.right !is n)
1818                             (n.right !is null && n.right.color == n.color.Red))
1823                 int r = recurse(n.right, path ~ "R");
1829                         "Node at path " ~ path ~ " has different number of black nodes on left and right paths"