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Design Analysis of Algorithm

Deletion In Red Black Tree

Case 1: x's sibling w is red

Case 1 (lines 5-8 of RB-DELETE-FIXUP and Figure 13.7(a)) occurs when node w, the sibling of node x, is red. Since w must have black children, we can switch the colors of w and p[x] and then perform a left-rotation on p[x] without violating any of the red-black properties. The new sibling of x, which is one of w's children prior to the rotation, is now black, and thus we have converted case 1 into case 2, 3, or 4.

Cases 2, 3, and 4 occur when node w is black; they are distinguished by the colors of w's children.

Case 2: x's sibling w is black, and both of w's children are black

In case 2 (lines 10-11 of RB-DELETE-FIXUP and Figure 13.7(b)), both of w's children are black. Since w is also black, we take one black off both x and w, leaving x with only one black and leaving w red. To compensate for removing one black from x and w, we would like to add an extra black to p[x], which was originally either red or black. We do so by repeating the while loop with p[x] as the new node x. Observe that if we enter case 2 through case 1, the new node x is red-and-black, since the original p[x] was red. Hence, the value c of the color attribute of the new node x is RED, and the loop terminates when it tests the loop condition. The new node x is then colored (singly) black in line 23.

Case 3: x's sibling w is black, w's left child is red, and w's right child is black

Case 3 (lines 13-16 and Figure 13.7(c)) occurs when w is black, its left child is red, and its right child is black. We can switch the colors of w and its left child left[w] and then perform a right rotation on w without violating any of the red-black properties. The new sibling w of x is now a black node with a red right child, and thus we have transformed case 3 into case 4.

Case 4: x's sibling w is black, and w's right child is red

Case 4 (lines 17-21 and Figure 13.7(d)) occurs when node x's sibling w is black and w's right child is red. By making some color changes and performing a left rotation on p[x], we can remove the extra black on x, making it singly black, without violating any of the red-black properties. Setting x to be the root causes the while loop to terminate when it tests the loop condition.

Analysis

What is the running time of RB-DELETE? Since the height of a red-black tree of n nodes is O(lg n), the total cost of the procedure without the call to RB-DELETE-FIXUP takes O(lg n) time. Within RB-DELETE-FIXUP, cases 1, 3, and 4 each terminate after performing a constant number of color changes and at most three rotations. Case 2 is the only case in which the while loop can be repeated, and then the pointer x moves up the tree at most O(lg n) times and no rotations are performed. Thus, the procedure RB-DELETE-FIXUP takes O(lg n) time and performs at most three rotations, and the overall time for RB-DELETE is therefore also O(lg n).

AVL trees

An AVL tree is a binary search tree that is height balanced: for each node x, the heights of the left and right subtrees of x differ by at most 1. To implement an AVL tree, we maintain an extra field in each node: h[x] is the height of node x. As for any other binary search tree T, we assume that root[T] points to the root node.

  1. Prove that an AVL tree with n nodes has height O(lg n). (Hint: Prove that in an AVL tree of height h, there are at least Fh nodes, where Fh is the hth Fibonacci number.)

  2. To insert into an AVL tree, a node is first placed in the appropriate place in binary search tree order. After this insertion, the tree may no longer be height balanced. Specifically, the heights of the left and right children of some node may differ by 2. Describe a procedure BALANCE(x), which takes a subtree rooted at x whose left and right children are height balanced and have heights that differ by at most 2, i.e., |h[right[x]] - h[left[x]]| ≤ 2, and alters the subtree rooted at x to be height balanced. (Hint: Use rotations.)

  3. Using part (b), describe a recursive procedure AVL-INSERT(x, z), which takes a node x within an AVL tree and a newly created node z (whose key has already been filled in), and adds z to the subtree rooted at x, maintaining the property that x is the root of an AVL tree. As in TREE-INSERT from Section 12.3, assume that key[z] has already been filled in and that left[z] = NIL and right[z] = NIL; also assume that h[z] = 0. Thus, to insert the node z into the AVL tree T, we call AVL-INSERT(root[T], z).

  4. Give an example of an n-node AVL tree in which an AVL-INSERT operation causes Ω(lg n) rotations to be performed.