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A tournament tree can be represented as a balanced binary tree by adding sentinels to the input lists (i.e. adding a member to the end of each list with a value of infinity) and by adding null lists (comprising only a sentinel) until the number of lists is a power of two. The balanced tree can be stored in a single array.
Else, recursively merge the first ⌊k/2⌋ lists and the final ⌈k/2⌉ lists, then binary merge these. When the input lists to this algorithm are ordered by length, shortest first, it requires fewer than n ⌈log k ⌉ comparisons, i.e., less than half the number used by the heap-based algorithm; in practice, it may be about as fast or slow ...
In part 2 a slightly more complex merge happens. The tree with the lower value (tree x) has a right child, so merge must be called again on the subtree rooted by tree x's right child and the other tree. After the merge with the subtree, the resulting tree is put back into tree x. The s-value of the right child (s=2) is now greater than the s ...
To merge two binomial trees of the same order, first compare the root key. Since 7>3, the black tree on the left (with root node 7) is attached to the grey tree on the right (with root node 3) as a subtree. The result is a tree of order 3. The operation of merging two heaps is used as a subroutine in most other operations. A basic subroutine ...
If the two trees are balanced, join simply creates a new node with left subtree t 1, root k and right subtree t 2. Suppose that t 1 is heavier (this "heavier" depends on the balancing scheme) than t 2 (the other case is symmetric). Join follows the right spine of t 1 until a node c which is balanced with t 2.
Join: The function Join is on two weight-balanced trees t 1 and t 2 and a key k and will return a tree containing all elements in t 1, t 2 as well as k. It requires k to be greater than all keys in t 1 and smaller than all keys in t 2. If the two trees have the balanced weight, Join simply create a new node with left subtree t 1, root k and ...
Next, c, d, and e are read. A one-node tree is created for each and a pointer to the corresponding tree is pushed onto the stack. Creating a one-node tree. Continuing, a '+' is read, and it merges the last two trees. Merging two trees. Now, a '*' is read. The last two tree pointers are popped and a new tree is formed with a '*' as the root ...
Insertion of a number in a 2–3 tree for 3 possible cases. If the target node is a 3-node whose parent is a 2-node, the key is inserted into the 3-node to create a temporary 4-node. In the illustration, the key 10 is inserted into the 2-node with 6 and 9. The middle key is 9, and is promoted to the parent 2-node.