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Assignment No: 6 Binary Trees Solution
In this assignment, you construct a binary tree from user input. Assume that the nodes of the tree store distinct keys. Your task is to convert the input tree in place to a binary search tree with respect to these key values. The only operations permitted are child swaps and rotations (explained in Part 3). You are allowed to read keys stored in the nodes, but you are not allowed to swap keys in two different nodes. You are also not allowed to copy external data (keys) to the nodes. Your conversion process must not create any new node. You need to achieve your objective by making a sequence of permitted operations (child swaps and rotations) on the input binary tree. The final BST you produce should be balanced. Your algorithm should run in O(n2) time, where n is the number of nodes in the tree.
Part 1: The tree data structure
Write a user-defined data type to store a node in binary trees. Each node should store an integer key, and two child pointers (left and right). If you choose, you may include a parent pointer in each node. Besides these, the nodes must not store any additional data.
Part 2: Construct the tree from user input
Write a function readtree to construct a binary tree from user input. The user first specifies the number n of nodes in the tree. The user then specifies the keys in a preorder fashion. In addition to the key for a node, the user also specifies the numbers of nodes in the left and the right subtrees of that node. The sample output illustrates the user input. Assume that it is the user’s duty to supply n distinct key values, and subtree counts consistent with a binary tree with n nodes. Your function may return a pointer to the root of the tree. For a reason that will be clear soon, the function may additionally create a dummy root, make the actual root the right child of the dummy root, and return a pointer to the dummy root. The key of the dummy root will never be consulted, so this field may be kept uninitialized, or set to −∞.
Part 3: Utility functions
Write a function to print the preorder listing of the keys in a binary tree, and another function to print the inorder listing of the keys in a binary tree. You can uniquely construct a tree from these two listings. You may write another function prntree that calls these two listing functions.
In later parts, we need two operations on the nodes of a binary tree: (i) swapping the two child pointers of a node, and (ii) left/right rotation at a node. These operations are explained in the figure below. Write three functions swapchild, lrotate and rrotate. Each rotate function should return a pointer to the new root of the rotated subtree. The parent of the old root should change its appropriate child pointer to point to the new root. This is the reason why it is helpful for the root to have a dummy parent.
p q
p p
q p
W U
U V V U
U V V W
Swap child nodes Rotation
Note that the rotate operations do not change the inorder listing of the tree, but a child swap operation does (unless made on empty subtrees).
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Part 4: Make the tree completely right-skew
By making a sequence of right rotations, make the tree completely right skew. Start at the root, and keep on applying right rotations there until the left subtree becomes empty. Then proceed to the right child, and repeat. This process takes O(n) time. Write a function makeskew for this part.
Part 5: Bubble sort the skew tree
After Part 4, your tree is essentially a linked list connected by the right-child links. For a general binary tree, this list is not sorted in general. Write a function bsort to bubble sort this list. Bubble sort requires swapping of data in consecutive locations. However, changing the key values of nodes is not permitted. A child swap, a right rotation, and finally another child swap is a sequence that achieves swapping of two nodes in the linked list. The following figure illustrates this process. The running time is O(n2).
3
3
3
3
5
5
5
5
6
6
2
2
2
Swap child at 6
2
Right rotate at 6
6
Swap child at 6
6
1
1
1
1
8
8
8
8
4
4
4
4
7
7
7
7
Part 6: Rebalance the skew tree
Finally, we want a balanced tree in the sense that at each node v, we must have |left(v)| − |right(v)| ∈ {0, 1}, where |u| is the number of nodes in the subtree rooted at u. Write a function rebalance that works as follows. After Part 5, the tree is fully right-skew. Locate the middle node in the list. Keep on applying left rotations at the root until that middle element reaches the root position. At this point, the left subtree of the root is fully left-skew, and the right subtree of the root is fully right-skew. Again, locate the middle nodes in the two subtrees. Keep on applying rotations (right rotations at the left child of the root, and left rotations at the right child of the root) until the middle nodes are placed in the desired locations. Recurse. This process takes O(n log n) time.
The MAIN() function
• Call readtree to construct the input binary tree T . Print T .
• Call makeskew on T . Print T .
• Call bsort on the right-skew tree T . Print T .
• Call rebalance on the sorted right-skew tree. Print T .
Submit a single C/C++ source file. Do not use global/static variables.
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Sample output
The sample output corresponds to the trees given below.
215
191
192
192
205
216
194
205
201
201
192
194
205
205
194
215
215
216
191
213
191
213
192
201
213
216
194
215
201
213
216
191
The initial binary tree
After right skewing
After bubble sort
After rebalancing
n = 8
205
4
3
192
1
2
215 0 0
216 0 1
201
0
0
194
1
1
191
0
0
213
0
0
+++ Initial
tree
---
Preorder listing
205
192
215
216
201
194
191
213
---
Inorder
listing
215
192
216
201
205
191
194
213
• Tree made skew
--- Preorder listing
215
192
216
201
205
191
194
213
--- Inorder listing
215
192
216
201
205
191
194
213
• Tree after sorting
--- Preorder listing
191
192
194
201
205
213
215
216
--- Inorder listing
191
192
194
201
205
213
215
216
• Tree after rebalancing
--- Preorder listing
205
194
192
191
201
215
213
216
--- Inorder listing
191
192
194
201
205
213
215
216
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