AVL tree

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AVL tree
Type Tree
Invented 1962
Invented by Georgy Adelson-Velsky and E. M. Landis
Time complexity
in big O notation
Average Worst case
Space O(n) O(n)
Search O(log n) O(log n)
Insert O(log n) O(log n)
Delete O(log n) O(log n)
Example AVL tree

In computer science, an AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. It was the first such data structure to be invented.[1] In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. Lookup, insertion, and deletion all take O(log n) time in both the average and worst cases, where n is the number of nodes in the tree prior to the operation. Insertions and deletions may require the tree to be rebalanced by one or more tree rotations.

The AVL tree is named after its two Soviet inventors, Georgy Adelson-Velsky and E. M. Landis, who published it in their 1962 paper "An algorithm for the organization of information".[2]

AVL trees are often compared with red-black trees because both support the same set of operations and take O(log n) time for the basic operations. For lookup-intensive applications, AVL trees are faster than red-black trees because they are more rigidly balanced.[3] Similar to red-black trees, AVL trees are height-balanced. Both are in general not weight-balanced nor μ-balanced for any \scriptstyle \mu\leq\tfrac12;[4] that is, sibling nodes can have hugely differing numbers of descendants.


Tree rotations

Basic operations of an AVL tree involve carrying out the same actions as would be carried out on an unbalanced binary search tree, but modifications are followed by zero or more operations called tree rotations, which help to restore the height balance of the subtrees.


Once a node has been found in a balanced tree, the next or previous nodes can be explored in amortized constant time. Some instances of exploring these "nearby" nodes require traversing up to 2×log(n) links (particularly when moving from the rightmost leaf of the root's left sub tree to the leftmost leaf of the root's right sub tree; in the example AVL tree, moving from node 14 to the next but one node 19 takes 4 steps). However, exploring all n nodes of the tree in this manner would use each link exactly twice: one traversal to enter the sub tree rooted at that node, another to leave that node's sub tree after having explored it. And since there are n−1 links in any tree, the amortized cost is found to be 2×(n−1)/n, or approximately 2.


Pictorial description of how rotations cause rebalancing tree, and then retracing one's steps toward the root updating the balance factor of the nodes. The numbered circles represent the nodes being balanced. The lettered triangles represent subtrees which are themselves balanced BSTs. A blue number next to a node denotes possible balance factors.

After inserting a node, it is necessary to check each of the node's ancestors for consistency with the rules of AVL. The balance factor is calculated as follows: balanceFactor = height(left subtree) - height(right subtree). For each node checked, if the balance factor remains −1, 0, or +1 then no rotations are necessary. However, if balance factor becomes less than -1 or greater than +1, the subtree rooted at this node is unbalanced. If insertions are performed serially, after each insertion, at most one of the following cases needs to be resolved to restore the entire tree to the rules of AVL.

Let us first assume the balance factor of a node L is 2 (as opposed to the other possible unbalanced value -2). This case is depicted in the left column of the illustration with L=5. We then look at the left subtree (the larger one) with root P. If this subtree does not lean to the right - i.e. P has balance factor 0 or 1 - we can rotate the whole tree to the right to get a balanced tree. This is labelled as the "Left Left Case" in the illustration with P=4. If the subtree does lean to the right - i.e. P has balance factor -1 - we first rotate the subtree to the left and end up the previous case. The second case is labelled as "Left Right Case" with P=5 in the illustration.

If the balance factor of the node L is -2 (this case is depicted in the right column of the illustration L=3) we can mirror the above algorithm. I.e. if the root P of the right subtree has balance factor 0 or -1 we can rotate the whole tree to the left to get a balanced tree. This is labelled as the "Right Right Case" in the illustration with P=4. If the root P of the right subtree has balance factor 1 we can rotate the subtree to the right to end up in the "Right Right Case".

The whole algorithm looks like this:

 if (balance_factor(L) == 2) { //The left column
   let P=left_child(L)
   if (balance_factor(P) == -1) { //The "Left Right Case"
      rotate_left(P) //reduce to "Left Left Case"
   //Left Left Case
 } else { // balance_factor(L) == -2, the right column
   let P=right_child(L)
   if (balance_factor(P) == 1) { //The "Right Left Case"
      rotate_right(P) //reduce to "Right Right Case"
   //Right Right Case

The names of the cases refer to the portion of the tree that is reduced in height.

In order to restore the balance factors of all nodes, first observe that all nodes requiring correction lie along the path used during the initial insertion. If the above procedure is applied to nodes along this path, starting from the bottom (i.e. the node furthest away from the root), then every node in the tree will again have a balance factor of -1, 0, or 1.


Let node X be the node with the value we need to delete, and let node Y be a node in the tree we need to find to take node X's place, and let node Z be the actual node we take out of the tree.

Steps to consider when deleting a node in an AVL tree are the following:

  1. If node X is a leaf or has only one child, skip to step 5. (node Z will be node X)
  2. Otherwise, determine node Y by finding the largest node in node X's left sub tree (in-order predecessor) or the smallest in its right sub tree (in-order successor).
  3. Replace node X with node Y (remember, tree structure doesn't change here, only the values). In this step, node X is essentially deleted when its internal values were overwritten with node Y's.
  4. Choose node Z to be the old node Y.
  5. Attach node Z's subtree to its parent (if it has a subtree). If node Z's parent is null, update root. (node Z is currently root)
  6. Delete node Z.
  7. Retrace the path back up the tree (starting with node Z's parent) to the root, adjusting the balance factors as needed.

As with all binary trees, a node's in-order successor is the left-most child of its right subtree, and a node's in-order predecessor is the right-most child of its left subtree. In either case, this node will have zero or one children. Delete it according to one of the two simpler cases above.

Deleting a node with two children from a binary search tree using the inorder predecessor (rightmost node in the left subtree, labelled 6).

In addition to the balancing described above for insertions, if the balance factor for the tree is 2 and that of the left subtree is 0, a right rotation must be performed on P. The mirror of this case is also necessary.

The retracing can stop if the balance factor becomes −1 or +1 indicating that the height of that subtree has remained unchanged. If the balance factor becomes 0 then the height of the subtree has decreased by one and the retracing needs to continue. If the balance factor becomes −2 or +2 then the subtree is unbalanced and needs to be rotated to fix it. If the rotation leaves the subtree's balance factor at 0 then the retracing towards the root must continue since the height of this subtree has decreased by one. This is in contrast to an insertion where a rotation resulting in a balance factor of 0 indicated that the subtree's height has remained unchanged.

The time required is O(log n) for lookup, plus a maximum of O(log n) rotations on the way back to the root, so the operation can be completed in O(log n) time.

Comparison to other structures[edit]

Both AVL trees and red-black trees are self-balancing binary search trees and they are very similar mathematically.[5] The operations to balance the trees are different, but both occur on the average in O(1) with maximum in O(log n). The real difference between the two is the limiting height. For a tree of size  n :

AVL trees are more rigidly balanced than red-black trees, leading to slower insertion and removal but faster retrieval.

See also[edit]


  1. ^ Robert Sedgewick, Algorithms, Addison-Wesley, 1983, ISBN 0-201-06672-6, page 199, chapter 15: Balanced Trees.
  2. ^ Georgy Adelson-Velsky, G.; E. M. Landis (1962). "An algorithm for the organization of information". Proceedings of the USSR Academy of Sciences 146: 263–266.  (Russian) English translation by Myron J. Ricci in Soviet Math. Doklady, 3:1259–1263, 1962.
  3. ^ Pfaff, Ben (June 2004). "Performance Analysis of BSTs in System Software" (PDF). Stanford University. 
  4. ^ AVL trees are not weight-balanced? (meaning: AVL trees are not μ-balanced?)
    Thereby: A Binary Tree is called \mu-balanced, with 0 \le\mu\leq\tfrac12, if for every node N, the inequality
    \tfrac12-\mu\le\tfrac{|N_l|}{|N|+1}\le \tfrac12+\mu
    holds and \mu is minimal with this property. |N| is the number of nodes below the tree with N as root (including the root) and N_l is the left child node of N.
  5. ^ In fact, each AVL tree can be colored red-black.
  6. ^ Burkhard, Walt (Spring 2012). "AVL Dictionary Data Type Implementation". Advanced Data Structures. La Jolla: A.S. Soft Reserves, UC San Diego. p. 103. 
  7. ^ Knuth, Donald E. (2000). Sorting and searching (2. ed., 6. printing, newly updated and rev. ed.). Boston [u.a.]: Addison-Wesley. p. 460. ISBN 0-201-89685-0. 
  8. ^ Proof of asymptotic bounds

Further reading[edit]

  • Donald Knuth. The Art of Computer Programming, Volume 3: Sorting and Searching, Third Edition. Addison-Wesley, 1997. ISBN 0-201-89685-0. Pages 458–475 of section 6.2.3: Balanced Trees.

External links[edit]