IPR Tree Example
Slides by Jagoda Walny
CPSC 335, Tutorial 02
Winter 2008
Initial IPR tree:
57
26
72
25
3
38
37
63
47
94
78
Internal path length:
4x4 + 4x3 + 2x2 + 1 = 33
Next step: insert 1 -->
Insert 1 and see if tree can be rebalanced
57
26
72
25
Start with parent
of inserted node.
3
38
37
63
47
94
78
1
Internal path length:
1x5 + 4x4 + 4x3 + 2x2 + 1 = 38
Next step: Find rebalancing case -->
Find rebalancing case
57
26
72
25
3
38
37
63
47
94
78
1
We can apply a single right rotation at node 25 since it
is ML (more nodes to the left), and the number of nodes
in the left subtree of node 3 is 1 and the number of nodes
in the right subtree of node 25 is 0, and 1 > 0.
Next step: apply single right rotation -->
Apply single right rotation
57
26
72
3
1
38
25
37
63
47
94
78
Internal path length:
5x4 + 4x3 + 2x2 + 1 = 37
Next step: insert 30 -->
Insert 30 and see if tree can be rebalanced
57
26
72
3
1
38
25
37
63
47
94
78
30
1. Single right rotation at node 38: cannot be applied
since the left subtree of 37 and the right subtree of 38
contain the same number of nodes (1).
2. Double rotation at node 26: cannot be applied since
the subtrees of 37 have fewer nodes (1) than the left
subtree of 26 (3).
Internal path length:
1x5 + 5x4 + 4x3 + 2x2 + 1 = 42
3. Double rotation at node 57: cannot be applied since
the subtrees of 38 contain fewer nodes (3) than the right
subtree of 57 (4).
So: tree cannot be rebalanced.
Next step: Insert 32 -->
Insert 32 and see if tree can be rebalanced
57
26
72
3
1
38
25
37
63
47
94
78
30
32
Internal path length:
1x6 + 1x5 + 5x4 + 4x3 + 2x2 + 1 = 48
Next step: Find rebalancing case -->
Find rebalancing case
57
26
72
3
1
38
25
37
63
47
94
78
30
32
We can apply a double rotation since the number of
nodes in the right subtree of node 30 is 1 and the
number of nodes in the right subtree of node 37 is 0, and
1 > 0.
Next step: Perform double rotation -->
Perform double rotation
57
26
72
3
1
38
32
25
30
63
47
94
78
37
Internal path length:
2x5 + 5x4 + 4x3 + 2x2 + 1 = 47
Next step: Insert 35 -->
Insert 35 and see if tree can be rebalanced
57
26
72
3
1
38
32
25
30
63
47
94
78
37
35
Internal path length:
1x6 + 2x5 + 5x4 + 4x3 + 2x2 + 1 = 53
Next step: Find rebalancing case -->
Find rebalancing case
57
26
72
3
1
38
32
25
30
63
47
94
78
37
35
We can apply a double rotation since the number of
nodes in the right subtree of node 32 is 2 and the
number of nodes in the right subtree of node 38 is 1, and
2 > 1.
Next step: Apply double rotation -->
Apply double rotation
57
26
72
3
1
37
25
32
30
63
38
35
94
78
47
Internal path length:
3x5 + 5x4 + 4x3 + 2x2 + 1 = 52
Next step: Notes -->
Notes
• When looking for possible rotations, always start at the parent of the just-inserted node, see if it meets
the criteria for any of the 4 cases of rotations, and if not, try its parent.
• If you have a choice between a single rotation and a double rotation, do a single rotation since it is
simpler.
• Note that after every rotation done in this example, the internal path length was reduced.
• Note that the results of this example do not differ from those of the AVL example.
• As an exercise, try inserting some more nodes into this tree.