How is Wikipedia's example of an unbalanced AVL tree really unbalanced?

Question

The image above is from "Wikipedia's entry on AVL trees" which Wikipedia indicates is unbalanced. How is this tree not balanced already? Here's a quote from the article:

The balance factor of a node is the height of its right subtree minus the height of its left subtree and a node with balance factor 1, 0, or -1 is considered balanced. A node with any other balance factor is considered unbalanced and requires rebalancing the tree. The balance factor is either stored directly at each node or computed from the heights of the subtrees.

Both the left and right subtrees have a height of 4. The right subtree of the left tree has a height of 3 which is still only 1 less than 4. Can someone explain what I'm missing?

Answer 1

To be balanced, every node in the tree must, either,

• have no children, (be a "leaf" node)
• Have two children.

• Or, if it has only one child, that child must be a leaf.

  In the chart you posted, 9, 54 & 76 violate the last rule.

Properly balanced, the tree would look like:

Root: 23
(23) -> 14 & 67
(14) -> 12 & 17
(12) -> 9
(17) -> 19
(67) -> 50 & 72
(50) -> 54
(72) -> 76

UPDATE: (after almost 9 years, I created a cool online chart for the graph at draw.io)

Answer 2

Node 76 is unbalanced, for example, because its right subtree is of height 0 and the left is of height 3.

Answer 3

Intuitively, it's because it's not as small as possible. e.g., 12 should be the parent of 9 and 14. As it is, 9 has no left sub-tree so it's out of balance. A tree is a hierarchical data structure so a rule like "balanced" often apply to every node and not just the root node.

You're correct the root node is balanced, but not all the nodes of the tree are.

Answer 4

Another way to look at this is that the height h of any node is given by:

h = 1 + max( left.height, right.height )

and a node is unbalanced whenever:

abs( left.height - right.height ) > 1

Looking at the tree above:

- Node 12 is a leaf node so its height = 1+max(0,0) = 1
- Node 14 has one child (12, on the left), so its height is = 1+max(1,0) = 2
- Node 9 has one child (14, on the right), so its height is = 1+max(0,2) = 3

To determine if node 9 is balanced or not we look at the height of its children:

- 9's left child is NULL, so 9.left.height = 0
- 9's right child (14) has height 2, so 9.right.height = 2

Now solve to show that node 9 is unbalanced:

9.unbalanced = abs( 9.left.height - 9.right.height ) > 1
9.unbalanced = abs( 0 - 2 ) > 1
9.unbalanced = abs( -2 ) > 1
9.unbalanced = 2 > 1
9.unbalanced = true

Similar calculations can be made for every other node.