Data Structures (CS 1520) Lecture 23 Name:

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1 ata Structures (S 152) Lecture 23 Name: 1. n VL ree is a special type of inary Search ree (S) that it is balanced. y balanced I mean that the of every s left and right subtrees differ by at most one. his is enough to guarantee that a VL tree with n s has a no worst than O(1.44 log 2 n). herefore, insertions, deletions, and search are worst case O( log 2 n ). n example of an VL tree with integer keys is shown below. he of each is shown ach VL-tree usually stores a balance factor in addition to its key and payload. he balance factor keeps track of the relative difference between its left and right subtrees, i.e., (left subtree) - (right subtree). a) Label each in the above VL tree with one of the following balance factors: if its left and right subtrees are the same 1 if its left subtree is one taller than its right subtree -1 if its right subtree is one taller than its left subtree b) We start a put operation by adding the item into the VL as a leaf just like we did for inary Search rees (Ss). dd the key 9 to the above tree. c) Identify the closest up the tree" from the inserted (9) that no longer satisfies the -balanced property of an VL tree. his is called the pivot. Label the pivot above. d) onsider the subtree whose root is the pivot. How could we rearrange this subtree to restore the VL balanced property? (raw the rearranged tree below) Lecture 23 Page 1

2 ata Structures (S 152) Lecture 23 Name: 2. ypically, the addition of a key into an VL requires the following steps: compare the key with the current tree s key (as we did in the _put function called by the put method in the S) to determine whether to recursively add the key into the left or right subtree add the key as a leaf as the base case(s) to the recursion recursively (updatealance method) adjust the balance factors of the s on the search path from the back up toward the root of the tree. If we encounter a pivot (as in question (c) above) we perform one or two rotations to restore the VL tree s -balanced property. or example, consider the previous example of adding 9 to the VL tree. efore the addition, the pivot (6) was already -1 ( tall right - right subtree had a one greater than its left subtree). fter inserting 9, the pivot s right subtree had a 2 more than its left subtree (balance factor -2) which violates the VL tree s -balance property. his problem is handled with a left rotation about the pivot as shown in the following generalized diagram: efore the addition: fter the addition, but before rotation: -1-2 Left at -1 Pivot Recursive updatealance method finds the pivot and calls the rebalance method to perform proper rotation(s) ('s balance factor was already adjusted before the pivot is found by the recursive updatealance method which moves toward the root) n fter left rotation at pivot: n a) ssuming the same initial VL tree (upper, left-hand of above diagram) if the would have increased the of (instead of ), would a left rotation about the have rebalanced the VL tree? Lecture 23 Page 2

3 ata Structures (S 152) Lecture 23 Name: b) efore the addition, if the pivot was already -1 (tall right) and if the is inserted into the left subtree of the pivot 's right child, then we must do two rotations to restore the VL-tree s -balance property. efore the addition: fter the addition, but before first rotation: Recursive updatealance finds the pivot -1-2 and calls rebalance method to perform rotation(s) 's & 's balance factors have already been adjusted before 1 the pivot was found 1st -1 Right at Node n - 2 n - 2 n - 2 fter the left rotation at pivot and balance factors adjusted correctly: fter right rotation at, but before left rotation at pivot: -2 1 n - 2 Left at Pivot n b) Suppose that the was added in instead of, then the same two rotations would restore the VL-tree s -balance property. However, what should the balance factors of s,, and be after the rotations? Lecture 23 Page 3

4 ata Structures (S 152) Lecture 23 Name: onsider the VLreeNode class that inherits and extends the reenode class to include balance factors. from tree_ import reenode class VLreeNode(reeNode): def init (self,key,val,left=none,right=none,parent=none, balanceactor=): reenode. init (self,key,val,left,right,parent) self.balanceactor = balanceactor Now let s consider the partial VLree class code that inherits from the inarysearchree class: from avl_tree_ import VLreeNode from binary_search_tree import inarysearchree class VLree(inarySearchree): def put(self,key,val): if self.root: self._put(key,val,self.root) self.root = VLreeNode(key,val) self.size = self.size + 1 def _put(self,key,val,currentnode): if key < currentnode.key: if currentnode.haslefthild(): self._put(key,val,currentnode.lefthild) currentnode.lefthild = VLreeNode(key,val,parent=currentNode) self.updatealance(currentnode.lefthild) elif key > currentnode.key: if currentnode.hasrighthild(): self._put(key,val,currentnode.righthild) currentnode.righthild = VLreeNode(key,val,parent=currentNode) self.updatealance(currentnode.righthild) currentnode.payload = val self._size -= 1 def updatealance(self,): if.balanceactor > 1 or.balanceactor < -1: self.rebalance() return if.parent!= None: if.islefthild():.parent.balanceactor += 1 elif.isrighthild():.parent.balanceactor -= 1 if.parent.balanceactor!= : self.updatealance(.parent) def rotateleft(self,rotroot): ## NO: You will complete rotateright in Lab Root = rotroot.righthild rotroot.righthild = Root.lefthild if Root.lefthild!= None: Root.lefthild.parent = rotroot Root.parent = rotroot.parent if rotroot.isroot(): self.root = Root if rotroot.islefthild(): rotroot.parent.lefthild = Root rotroot.parent.righthild = Root Root.lefthild = rotroot rotroot.parent = Root rotroot.balanceactor = rotroot.balanceactor min(root.balanceactor, ) Root.balanceactor = Root.balanceactor max(rotroot.balanceactor, ) def rebalance(self,): if.balanceactor < : if.righthild.balanceactor > : self.rotateright(.righthild) self.rotateleft() self.rotateleft() elif.balanceactor > : if.lefthild.balanceactor < : self.rotateleft(.lefthild) self.rotateright() self.rotateright() Lecture 23 Page 4

5 ata Structures (S 152) Lecture 23 Name: c) race the code for myvl.put(9,none)by updating the below diagram: myvl VLree object size root onsider balance factor formulas for rotateleft. We know: al() = h - h and oldal() = h - (1+max(h, h )) al() = 1+ max(h, h ) - h and oldal() = h - h efore left rotation: rotroot fter left rotation at pivot: Root -min(-x, -y) = y max(x, y) = y -y -x x y -max(-x, -y) = x min(x, y) = x h Left at Pivot h h Root h rotroot d) onsider: al() - oldal() h h onsider: al() - oldal() al() - oldal() = h - h - h + (1+max(h, h )) al() - oldal() = 1 + max(h, h ) - h al() - oldal() = 1 + max(h, h ) - h al() = oldal() max(h - h, h - h ) al() = oldal() max(, -oldal()) al() = oldal() min(, oldal()), so rotroot.balanceactor = rotroot.balanceactor min(root.balanceactor, ) Lecture 23 Page 5

6 ata Structures (S 152) Lecture 23 Name: 3. omplete the below figure which is a mirror image to the figure on page 2, i.e., inserting into the pivot s left child s left subtree. Include correct balance factors after the rotation. efore the insertion: fter the insertion, but before rotation: Right at Pivot n fter right rotation at pivot: b) omplete the below figure which is a mirror image to the figure on page 3, i.e., inserting into the pivot s left child s right subtree. Include correct balance factors after the rotation. efore the insertion: fter the insertion, but before first rotation: n - 2 n - 2 n - 2 fter the right rotation at pivot and balance factors adjusted correctly: fter left rotation at, but before right rotation at pivot: Lecture 23 Page 6