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If the tree is not empty, then we go down the root, and recursively go down the tree searching for the location to insert the new node. This traversal is guided by the comparison function. In this case, the node always replaces a NULL reference (left or right) of an external node in the tree i.e., the node is either made a left-child or a right ...
But I agree, it would be interesting to state an algorithm how to do it. A simple one might be to create an empty tree, then walk the original unbalanced tree in-order, and insert each node into the new tree, using the AVL-insert. That should take O(n*log n). --Allefant 14:24, 24 November 2006 (UTC)
Weak AVL rule: all rank differences are 1 or 2, and all leaf nodes have rank 0. Note that weak AVL tree generalizes the AVL tree by allowing for 2,2 type node. A simple proof shows that a weak AVL tree can be colored in a way that represents a red-black tree. So in a sense, weak AVL tree combines the properties of AVL tree and red-black tree.
For example, leaf nodes by definition have no descendants, so given only a pointer to a leaf node no other node can be reached. A threaded tree adds extra information in some or all nodes, so that for any given single node the "next" node can be found quickly, allowing tree traversal without recursion and the extra storage (proportional to the ...
Join follows the right spine of t 1 until a node c which is balanced with t 2. At this point a new node with left child c, root k and right child t 2 is created to replace c. The new node may invalidate the balancing invariant. This can be fixed with rotations. The following is the join algorithms on different balancing schemes. The join ...
A node which transfers control to a node A is called an immediate predecessor of A. The dominance frontier of node A is the set of nodes B where A does not strictly dominate B, but does dominate some immediate predecessor of B. These are the points at which multiple control paths merge back together into a single path.
To turn a regular search tree into an order statistic tree, the nodes of the tree need to store one additional value, which is the size of the subtree rooted at that node (i.e., the number of nodes below it). All operations that modify the tree must adjust this information to preserve the invariant that size[x] = size[left[x]] + size[right[x]] + 1
The user can search for elements in an associative array, and delete elements from the array. The following shows how multi-dimensional associative arrays can be simulated in standard AWK using concatenation and the built-in string-separator variable SUBSEP: