A Binary Search Tree (BST) is a binary tree in which each vertex has only up to 2 children that satisfies BST property: All vertices in the left subtree of a vertex must hold a value smaller than its own and all vertices in the right subtree of a vertex must hold a value larger than its own (we have assumption that all values are distinct integers in this visualization and small tweak is needed to cater for duplicates/non integer). Try clicking for a sample animation on searching a random value ∈ [1..99] in the random BST above.
An Adelson-Velskii Landis (AVL) tree is a self-balancing BST that maintains it's height to be O(log N) when having N vertices in the AVL tree.
To toggle between the standard Binary Search Tree and the AVL Tree (only different behavior during Insertion and Removal of an Integer), select the respective header.
We also have URL shortcut to quickly access the AVL Tree mode, which is https://visualgo.net/en/avl (you can change the 'en' to your two characters preferred language - if available).
BST (and especially balanced BST like AVL Tree) is an efficient data structure to implement a certain kind of Table (or Map) Abstract Data Type (ADT).
A Table ADT must support at least the following three operations as efficient as possible:
Reference: See similar slide in Hash Table e-Lecture.
此e-Lecture的目标是引入BST然后平衡BST（AVL树）数据结构，以便我们可以实现基本的表ADT操作：搜索（v），插入（v），删除（v）和一些 其他Table ADT操作 - 参见下一张幻灯片 - 在O（log N）时间内 - 远小于N.
PS：一些经验丰富的读者可能会注意到另一种数据结构 ∃ 可以在更快的时间内实现三种基本的表ADT操作，但请继续阅读......
| N | ≈ 1 000 | ≈ 1 000 000 | ≈ 1 000 000 000 |
| log N | 10 | Only 20 :O | Only 30 :O:O |
On top of the basic three, there are a few other possible Table ADT operations:
Discussion: What are the best possible implementation for the first three additional operations if we are limited to use [sorted|unsorted] array/vector?
可用于实现表ADT的更简单的数据结构是链接列表。 Quiz: Can we perform all basic three Table ADT operations: Search(v)/Insert(v)/Remove(v) efficiently (read: faster than O(N)) using Linked List?
讨论：为什么？
可用于实现表ADT的另一种数据结构是哈希表。 它具有非常快的搜索（v），插入（v）和删除（v）性能（所有都在预期的O(1) 时间内）。 Quiz: So what is the point of learning this BST module if Hash Table can do the crucial Table ADT operations in unlikely-to-be-beaten expected O(1) time?
讨论上面的答案！ 提示：回到之前的4张幻灯片。
We will now introduce BST data structure. See the visualization of an example BST above!
Root vertex does not have a parent. There can only be one root vertex in a BST. Leaf vertex does not have any child. There can be more than one leaf vertex in a BST. Vertices that are not leaf are called the internal vertices. Sometimes root vertex is not included as part of the definition of internal vertex as the root of a BST with only one vertex can actually fit into the definition of a leaf too.
In the example above, vertex 15 is the root vertex, vertex {5, 7, 50} are the leaves, vertex {4, 6, 15 (also the root), 23, 71} are the internal vertices.
There are a few other BST (Query) operations that have not been visualized in VisuAlgo:
The details of these two operations are currently hidden for pedagogical purpose in a certain NUS module.
Because of the way data (distinct integers for this visualization) is organised inside a BST, we can binary search for an integer v efficiently (hence the name of Binary Search Tree).
First, we set the current vertex = root and then check if the current vertex is smaller/equal/larger than integer v that we are searching for. We then go to the right subtree/stop/go the left subtree, respectively. We keep doing this until we either find the required vertex or we don't.
On the example BST above, try clicking (found after 2 comparisons), (found after 3 comparisons), (not found after 3 comparisons).
We can perform an Inorder Traversal of this BST to obtain a list of sorted integers inside this BST (in fact, if we 'flatten' the BST into one line, we will see that the vertices are ordered from smallest/leftmost to largest/rightmost).
Inorder Traversal is a recursive method whereby we visit the left subtree first, exhausts all items in the left subtree, visit the current root, before exploring the right subtree and all items in the right subtree. Without further ado, let's try to see it in action on the example BST above.
无论BST的高度如何，Inorder Traversal都以O（N）运行。
讨论：为什么？
PS：有些人调用N个无序整数插入O（N log N）中的BST，然后执行O（N）Inorder Traversal作为'BST sort'。 它很少使用，因为有几种比这更容易使用（基于比较）的排序算法。
We can insert a new integer into BST by doing similar operation as Search(v). But this time, instead of reporting that the new integer is not found, we create a new vertex in the insertion point and put the new integer there. Try on the example above.
Quiz: Inserting integers [1,10,2,9,3,8,4,7,5,6] one by one in that order into an initially empty BST will result in a BST of height:
The first case is the easiest: Vertex v is currently one of the leaf vertex of the BST.
Deletion of a leaf vertex is very easy: We just remove that leaf vertex — try on the example BST above (second click onwards after the first removal will do nothing — please refresh this page or go to another slide and return to this slide instead).
This part is clearly O(1) — on top of the earlier O(h) search-like effort.
The second case is also not that hard: Vertex v is an (internal/root) vertex of the BST and it has exactly one child. Removing v without doing anything else will disconnect the BST.
Deletion of a vertex with one child is not that hard: We connect that vertex's only child with that vertex's parent — try on the example BST above (second click onwards after the first removal will do nothing — please refresh this page or go to another slide and return to this slide instead).
This part is also clearly O(1) — on top of the earlier O(h) search-like effort.
The third case is the most complex among the three: Vertex v is an (internal/root) vertex of the BST and it has exactly two children. Removing v without doing anything else will disconnect the BST.
Deletion of a vertex with two children is as follow: We replace that vertex with its successor, and then delete its duplicated successor in its right subtree — try on the example BST above (second click onwards after the first removal will do nothing — please refresh this page or go to another slide and return to this slide instead).
This part requires O(h) due to the need to find the successor vertex — on top of the earlier O(h) search-like effort.
本案例3值得进一步讨论：
Remove(v) runs in O(h) where h is the height of the BST. Removal case 3 (deletion of a vertex with two children is the 'heaviest' but it is not more than O(h)).
As you should have fully understand by now, h can be as tall as O(N) in a normal BST as shown in the random 'skewed right' example above. If we call , i.e. we remove the current max integer, we will go from root down to the last leaf in O(N) time before removing it — not efficient.
我们正在解释这个BST模块。 到目前为止，我们注意到许多基本的表格ADT操作在O（h）中运行，而h可以像N-1边缘一样高，就像所示的“向左倾斜”一样 - 效率低下:( ...
那么，有没有办法让我们的BST“不那么高”？
PS：如果你想研究如何在真实程序中实现这些基本的BST操作，你可以下载这个BSTDemo.cpp。
此时，我们建议您按[Esc]或单击此e-Lecture幻灯片右下角的X按钮进入“探索模式”并自行尝试各种BST操作，以加强您对这种多功能数据结构的理解。
当您准备继续解释平衡BST（我们使用AVL树作为示例）时，再次按[Esc]或从右上角的下拉菜单中将模式切换回“电子演讲模式”。 然后，使用幻灯片选择器下拉列表this slide 12-1恢复。
先前是:
v.height = -1 (if v is an empty tree)因此，BST的高度是: root.height.
v.height = max(v.left.height, v.right.height) + 1 (otherwise)
在上面的例子BST上, height(11) = height(32) = height(50) = height(72) = height(99) = 0 (所有都是叶子). height(29) = 1因为有1个边缘将它连接到它唯一的叶子32上。
Quiz: What are the values of height(20), height(65), and height(41) on the BST above?
height(41) = 4If we have N elements/items/keys in our BST, the lower bound height h > log2 N if we can somehow insert the N elements in perfect order so that the BST is perfectly balanced.
See the example shown above for N = 15 (a perfect BST which is rarely achievable in real life — try inserting any other integer and it will not be perfect anymore).
N ≤ 1 + 2 + 4 + ... + 2h
N ≤ 20 + 21 + 22 + … + 2h
N < 2h+1 (几何级数之和)
log2 N < log2 2h+1
log2 N < (h+1) * log2 2 (log2 2 is 1)
h > (log2 N)-1 (代数操作)
h > log2 N
If we have N elements/items/keys in our BST, the upper bound height h < N if we insert the elements in ascending order (to get skewed right BST as shown above).
The height of such BST is h = N-1, so we have h < N.
Discussion: Do you know how to get skewed left BST instead?
We have seen that most BST operations are in O(h) and combining the lower and upper bounds of h, we have log2 N < h < N.
There is a dramatic difference between log2 N and N and we have seen from the discussion of the lower bound that getting perfect BST (at all times) is near impossible...
So can we have BST that has height closer to log2 N, i.e. c * log2 N, for a small constant factor c? If we can, then BST operations that run in O(h) actually run in O(log N)...
Introducing AVL Tree, invented by two Russian (Soviet) inventors: Georgy Adelson-Velskii and Evgenii Landis, back in 1962.
In AVL Tree, we will later see that its height h < 2 * log N (tighter analysis exist, but we will use easier analysis in VisuAlgo where c = 2). Therefore, most AVL Tree operations run in O(log N) time — efficient.
Insert(v) and Remove(v) update operations may change the height h of the AVL Tree, but we will see rotation operation(s) to maintain the AVL Tree height to be low.
To have efficient performance, we shall not maintain height(v) attribute via the O(N) recursive method every time there is an update (Insert(v)/Remove(v)) operation.
Instead, we compute O(1): x.height = max(x.left.height, x.right.height) + 1 at the back of our Insert(v)/Remove(v) operation as only the height of vertices along the insertion/removal path may be affected. Thus, only O(h) vertices may change its height(v) attribute and in AVL Tree, h < 2 * log N.
Try on the example AVL Tree (ignore the resulting rotation for now, we will come back to it in the next few slides). Notice that only a few vertices along the insertion path: {41,20,29,32} increases their height by +1 and all other vertices will have their heights unchanged.
Adelson-Velskii and Landis claim that an AVL Tree (a height-balanced BST that satisfies AVL Tree invariant) with N vertices has height h < 2 * log2 N.
The proof relies on the concept of minimum-size AVL Tree of a certain height h.
Let Nh be the minimum number of vertices in a height-balanced AVL Tree of height h.
The first few values of Nh are N0 = 1 (a single root vertex), N1 = 2 (a root vertex with either one left child or one right child only), N2 = 4, N3 = 7, N4 = 12, N5 = 20 (see the background picture), and so on (see the next two slides).
我们知道，对于 N 个顶点的任何其他AVL树（不一定是最小尺寸的顶点），N ≥ Nh。
在背景图片中，我们有N5 = 20个顶点，但我们知道在我们有一个高度为h = 5的完美二叉树之前，我们可以再挤压43个顶点（最多N = 63）。
Nh = 1 + Nh-1 + Nh-2 (高度为h的最小尺寸AVL树的公式)
Nh > 1 + 2*Nh-2 (as Nh-1 > Nh-2)
Nh > 2*Nh-2 (obviously)
Nh > 4*Nh-4 (递归)
Nh > 8*Nh-6 (另一个递归步骤)
... (假设初始h是偶数，我们只能这样做h/2次)
Nh > 2h/2*N0 (我们到了base case)
Nh > 2h/2 (as N0 = 1)
N ≥ Nh > 2h/2 (结合前两张幻灯片)
N > 2h/2
log2(N) > log2(2h/2) (两张幻灯片都是log2 )
log2(N) > h/2 (等式化简)
2 * log2(N) > h or h < 2 * log2(N)
h = O(log(N)) (最后结论)
Look at the example BST again. See that all vertices are height-balanced, an AVL Tree.
To quickly detect if a vertex v is height balanced or not, we modify the AVL Tree invariant (that has absolute function inside) into: bf(v) = v.left.height - v.right.height.
Now try on the example AVL Tree again. A few vertices along the insertion path: {41,20,29,32} increases their height by +1. Vertices {29,20} will no longer be height-balanced after this insertion (and will be rotated later — discussed in the next few slides), i.e. bf(29) = -2 and bf(20) = -2 too. We need to restore the balance.
见上图。 在左侧图片上调用rotateRight(Q) 将生成正确的图片。 在右图上调用rotateLeft(P)将再次产生左图。
只有当T有一个左/右子时，才能调用rotateRight(T)/rotateLeft(T) 。
Tree Rotation保留BST属性。 在旋转之前，P≤B≤Q。旋转之后，请注意以B为根的子树（如果存在）更改父级，但P≤B≤Q不会更改。
BSTVertex rotateLeft(BSTVertex T) // pre-req: T.right != null
BSTVertex w = T.right // rotateRight is the mirror copy of this
w.parent = T.parent // this method is hard to get right for newbie
T.parent = w
T.right = w.left
if (w.left != null) w.left.parent = T
w.left = T
// update the height of T and then w here
return w
基本上，只有这四个不平衡的情况。 我们使用Tree Rotation(旋转树)来处理它们中的每一个。
Discussion: Is there other tree rotation cases for Insert(v) operation of AVL Tree?
与AVL树中的Insert（v）相比的主要区别在于，我们可能会多次触发四种可能的重新平衡情况中的一种，但不会超过h = O（log N）次：O，在上面的示例中尝试 看到两个连锁反应rotateRight（6）然后rotateRight（16）+ rotateLeft（8）组合。
We will end this module with a few more interesting things about BST and balanced BST (especially AVL Tree).
有关此数据结构的一些有趣的问题，请在BST/AVL培训模块上练习（无需登录）。
但是，对于注册用户，您应该登录然后转到主培训页面以正式清除此模块，此类成就将记录在您的用户帐户中。
We also have a few programming problems that somewhat requires the usage of this balanced BST (like AVL Tree) data structure: Kattis - compoundwords and Kattis - baconeggsandspam.
Try them to consolidate and improve your understanding about this data structure. You are allowed to use C++ STL map/set, Java TreeMap/TreeSet, or OCaml Map/Set if that simplifies your implementation (Note that Python doesn't have built-in bBST implementation).