all leaves are on the bottom level. all internal nodes (except perhaps the root node) have at least...

All leaves are on the bottom level.

All internal nodes (except perhaps the root node) have at least ceil(m / 2) (nonempty) children.

The root node can have as few as 2 children if it is an internal node, and can obviously have no children if the root node is a leaf (that is, the whole tree consists only of the root node).

Each leaf node (other than the root node if it is a leaf) must contain at least ceil(m / 2) - 1 keys.

A B-tree of order m is a multiway search tree of order m such that: Note that ceil(x) is the so-called ceiling function. It's value is the smallest integer that is greater than or equal to x. Thus ceil(3) = 3, ceil(3.35) = 4, ceil(1.98) = 2, ceil(5.01) = 6, ceil(7) = 7, etc. A B-tree is a fairly well-balanced tree by virtue of the fact that all leaf nodes must be at the bottom. Condition (2) tries to keep the tree fairly bushy by insisting that each node have at least half the maximum number of children. This causes the tree to "fan out" so that the path from root to leaf is very short even in a tree that contains a lot of data.

Let's work our way through an example similar to that given by Kruse. Insert the following letters into what is originally an empty B-tree of order 5: C N G A H E K Q M F W L T Z D P R X Y S Order 5 means that a node can have a maximum of 5 children and 4 keys. All nodes other than the root must have a minimum of 2 keys. The first 4 letters get inserted into the same node, resulting in this picture:

When we try to insert the H, we find no room in this node, so we split it into 2 nodes, moving the median item G up into a new root node. Note that in practice we just leave the A and C in the current node and place the H and N into a new node to the right of the old one.

Inserting E, K, and Q proceeds without requiring any splits:

Inserting M requires a split. Note that M happens to be the median key and so is moved up into the parent node.

The letters F, W, L, and T are then added without needing any split.

When Z is added, the rightmost leaf must be split. The median item T is moved up into the parent node. Note that by moving up the median key, the tree is kept fairly balanced, with 2 keys in each of the resulting nodes.

The insertion of D causes the leftmost leaf to be split. D happens to be the median key and so is the one moved up into the parent node. The letters P, R, X, and Y are then added without any need of splitting:

Finally, when S is added, the node with N, P, Q, and R splits, sending the median Q up to the parent. However, the parent node is full, so it splits, sending the median M up to form a new root node. Note how the 3 pointers from the old parent node stay in the revised node that contains D and G.

Deleting an Item

In the B-tree as we left it at the end of the last section, delete H. Of course, we first do a lookup to find H. Since H is in a leaf and the leaf has more than the minimum number of keys, this is easy. We move the K over where the H had been and the L over where the K had been. This gives

Next, delete the T. Since T is not in a leaf, we find its successor (the next item in ascending order), which happens to be W, and move W up to replace the T. That way, what we really have to do is to delete W from the leaf, which we already know how to do, since this leaf has extra keys. In ALL cases we reduce deletion to a deletion in a leaf, by using this method

Next, delete R. Although R is in a leaf, this leaf does not have an extra key; the deletion results in a node with only one key, which is not acceptable for a B-tree of order 5. If the sibling node to the immediate left or right has an extra key, we can then borrow a key from the parent and move a key up from this sibling. In our specific case, the sibling to the right has an extra key. So, the successor W of S (the last key in the node where the deletion occurred), is moved down from the parent, and the X is moved up. (Of course, the S is moved over so that the W can be inserted in its proper place.)

Finally, let's delete E. This one causes lots of problems. Although E is in a leaf, the leaf has no extra keys, nor do the siblings to the immediate right or left. In such a case the leaf has to be combined with one of these two siblings. This includes moving down the parent's key that was between those of these two leaves. In our example, let's combine the leaf containing F with the leaf containing A C. We also move down the D.

We begin by finding the immediate successor, which would be D, and move the D up to replace the C. However, this leaves us with a node with too few keys.

Topological Sort

• Introduction.

• Definition of Topological Sort.

• Topological Sort is Not Unique.

• Topological Sort Algorithm.

• An Example.

• Implementation.

• Review Questions.

Introduction• There are many problems involving a set of tasks in which

some of the tasks must be done before others.

• For example, consider the problem of taking a course only after taking its prerequisites.

• Is there any systematic way of linearly arranging the courses in the order that they should be taken?

Yes! - Topological sort.

Definition of Topological Sort• Topological sort is a method of arranging the vertices in a directed acyclic

graph (DAG), as a sequence, such that no vertex appear in the sequence before its predecessor.

• The graph in (a) can be topologically sorted as in (b)

(a) (b)

Topological Sort is not unique

• Topological sort is not unique.

• The following are all topological sort of the graph below:

s1 = {a, b, c, d, e, f, g, h, i}

s2 = {a, c, b, f, e, d, h, g, i}

s3 = {a, b, d, c, e, g, f, h, i}

s4 = {a, c, f, b, e, h, d, g, i}etc.

Topological Sort Algorithm• One way to find a topological sort is to consider in-degrees of the vertices.

• The first vertex must have in-degree zero -- every DAG must have at least one vertex with in-degree zero.

• The Topological sort algorithm is:int topologicalOrderTraversal( ){ int numVisitedVertices = 0; while(there are more vertices to be visited){ if(there is no vertex with in-degree 0) break; else{

select a vertex v that has in-degree 0; visit v; numVisitedVertices++; delete v and all its emanating edges;

return numVisitedVertices;}

Topological Sort Example• Demonstrating Topological Sort.

1 2 3 0 2

1 0 2 2 0

D G A B F H J E I C

Implementation of Topological Sort• The algorithm is implemented as a traversal method that visits the

vertices in a topological sort order.

• An array of length |V| is used to record the in-degrees of the vertices. Hence no need to remove vertices or edges.

• A priority queue is used to keep track of vertices with in-degree zero that are not yet visited.

public int topologicalOrderTraversal(Visitor visitor){ int numVerticesVisited = 0; int[] inDegree = new int[numberOfVertices]; for(int i = 0; i < numberOfVertices; i++) inDegree[i] = 0;

Iterator p = getEdges(); while (p.hasNext()) { Edge edge = (Edge) p.next(); Vertex to = edge.getToVertex(); inDegree[getIndex(to)]++; }

Implementation of Topological Sort BinaryHeap queue = new BinaryHeap(numberOfVertices); p = getVertices(); while(p.hasNext()){ Vertex v = (Vertex)p.next(); if(inDegree[getIndex(v)] == 0) queue.enqueue(v); } while(!queue.isEmpty() && !visitor.isDone()){ Vertex v = (Vertex)queue.dequeueMin(); visitor.visit(v); numVerticesVisited++; p = v.getSuccessors(); while (p.hasNext()){ Vertex to = (Vertex) p.next(); if(--inDegree[getIndex(to)] == 0) queue.enqueue(to); } } return numVerticesVisited;}

Review Questions

1. List the order in which the nodes of the directed graph GB are visited by topological order traversal that starts from vertex a.

2. What kind of DAG has a unique topological sort?

3. Generate a directed graph using the required courses for your major. Now apply topological sort on the directed graph you obtained.

What is a Graph?

• A graph G = (V,E) is composed of:V: set of verticesE: set of edges connecting the vertices in V

• An edge e = (u,v) is a pair of vertices• Example:

V= {a,b,c,d,e}

E= {(a,b),(a,c),(a,d),(b,e),(c,d),(c,e),(d,e)}

Applications

• electronic circuits

• networks (roads, flights, communications)

Terminology: Adjacent and Incident

• If (v0, v1) is an edge in an undirected graph, – v0 and v1 are adjacent– The edge (v0, v1) is incident on vertices v0 and v1

• If <v0, v1> is an edge in a directed graph– v0 is adjacent to v1, and v1 is adjacent from v0

– The edge <v0, v1> is incident on v0 and v1

The degree of a vertex is the number of edges incident to that vertexFor directed graph,

the in-degree of a vertex v is the number of edgesthat have v as the headthe out-degree of a vertex v is the number of edgesthat have v as the tailif di is the degree of a vertex i in a graph G with n vertices and e edges, the number of edges is

( ) /0

Terminology:Degree of a Vertex

Why? Since adjacent vertices each count the adjoining edge, it will be counted twice

3 4 5 6

1 1 1 1

directed graph

in-degreeout-degree

in:1, out: 1

in: 1, out: 2

in: 1, out: 0

Examples

Terminology:Path

• path: sequence of vertices v1,v2,. . .vk such that consecutive vertices vi and vi+1 are adjacent.

a b e d c b e d c

More Terminology• simple path: no repeated vertices

• cycle: simple path, except that the last vertex is the same as the first vertex

a c d a

Even More Terminology

• subgraph: subset of vertices and edges forming a graph• connected component: maximal connected subgraph. E.g., the graph below

has 3 connected components.

connected not connected

•connected graph: any two vertices are connected by some path

3 (i) (ii) (iii) (iv) (a) Some of the subgraph of G1

(i) (ii) (iii) (iv) (b) Some of the subgraph of G3

Subgraphs Examples

More…

• tree - connected graph without cycles• forest - collection of trees

forest

Directed vs. Undirected Graph

• An undirected graph is one in which the pair of vertices in a edge is unordered, (v0, v1) = (v1,v0)

• A directed graph is one in which each edge is a directed pair of vertices, <v0, v1> != <v1,v0>

tail head

Graph Representations

Adjacency MatrixAdjacency Lists

Adjacency Matrix

Let G=(V,E) be a graph with n vertices.The adjacency matrix of G is a two-dimensional n by n array, say adj_matIf the edge (vi, vj) is in E(G), adj_mat[i][j]=1If there is no such edge in E(G), adj_mat[i][j]=0The adjacency matrix for an undirected graph is symmetric; the adjacency matrix for a digraph need not be symmetric

Examples for Adjacency Matrix

symmetric

undirected: n2/2directed: n2

01234567

1 2 30 2 30 1 30 1 2

1 20 30 31 254 65 76G4

An undirected graph with n vertices and e edges ==> n head nodes and 2e list nodes

Depth-First Search 38

DFS : Depth-First Search

DFS is another popular search strategy. It can do certain things that BFS cannot do. We will discuss some of

these algorithms in COMP 271 (so you cannot get rid of DFS after COMP171).

DFS idea :

Whenever we visit a vertex v from another vertex u, we recursively visit a neighbor of v that has not been visited before until all neighbors of v have been visited. Then we backtrack (return) to u.

Algorithm

Flag all vertices as not visited

Visit v, and mark v as visited.

For each unvisited neighbor. make a recursive call RDFS(w).

Example

0Adjacency List

source

Visited Table (T/F)

Initialize visitedtable (all empty F)

Initialize Pred to -1

Example

0Adjacency List

source

Visited Table (T/F)

Mark 2 as visited

RDFS( 2 )recursive call RDFS(8)

visit sequence= {2}

Example

0 Adjacency List

source

Visited Table (T/F)

Mark 8 as visited

RDFS( 2 ) RDFS(8)

recursive callRDFS(0)

Recursivecalls

visit sequence= {2, 8}

Example

0 Adjacency List

source

Visited Table (T/F)

Mark 0 as visited

RDFS( 2 ) RDFS(8)

RDFS(0) -> no unvisited neighbor, return to (backtrack) RDFS(8)

Recursivecalls

visit sequence= {2, 8, 0}

Example

0 Adjacency List

source

Visited Table (T/F)

RDFS( 2 ) RDFS(8)

Recursivecalls

Backtrack to 8

visit sequence= {2, 8, 0}

Example

0 Adjacency List

source

Visited Table (T/F)

Mark 9 as visited

RDFS( 2 ) RDFS(8)

RDFS(9) recursive callRDFS(1)

Recursivecalls

visit sequence= {2, 8, 0, 9}

Example

0 Adjacency List

source

Visited Table (T/F)

Mark 1 as visited

RDFS( 2 ) RDFS(8)

RDFS(9) RDFS(1)

Recursivecalls

visit sequence= {2, 8, 0, 9, 1}

Example

0 Adjacency List

source

Visited Table (T/F)

Mark 3 as visited

PredRDFS( 2 ) RDFS(8)

RDFS(9) RDFS(1)

Recursivecalls

visit sequence= {2, 8, 0, 9, 1, 3}

RDFS( 2 ) RDFS(8)

RDFS(9) RDFS(1)

RDFS(3) RDFS(4) STOP all of 4’s neighbors have been visited backtrack (return back) to call RDFS(3)

Example

0 Adjacency List

source

Visited Table (T/F)

Mark 4 as visited

Recursivecalls

visit sequence= {2, 8, 0, 9, 1, 3, 4}

Example

0 Adjacency List

source

Visited Table (T/F)

RDFS(9) RDFS(1)

Recursivecalls

Backtrack to 3visit sequence= {2, 8, 0, 9, 1, 3, 4}

Example

0 Adjacency List

source

Visited Table (T/F)

RDFS(9) RDFS(1)

RDFS(3) RDFS(5) 3 is visited, recursive callRDFS(6)

Recursivecalls

Mark 5 as visited

visit sequence= {2, 8, 0, 9, 1, 3, 4, 5}

Example

0 Adjacency List

source

Visited Table (T/F)

RDFS( 2 ) RDFS(8)

RDFS(9) RDFS(1)

RDFS(3) RDFS(5) RDFS(6) recursive call RDFS(7)

Recursivecalls

Mark 6 as visited

visit sequence= {2, 8, 0, 9, 1, 3, 4, 5, 6}

Example

0 Adjacency List

source

Visited Table (T/F)

RDFS( 2 ) RDFS(8)

RDFS(9) RDFS(1)

RDFS(3) RDFS(5) RDFS(6)

RDFS(7)

Recursivecalls

Mark 7 as visited

visit sequence= {2, 8, 0, 9, 1, 3, 4, 5, 6, 7}

Example

0 Adjacency List

source

Visited Table (T/F)

RDFS( 2 ) RDFS(8)

RDFS(9) RDFS(1)

RDFS(3) RDFS(5) RDFS(6)

RDFS(7) no recursive call

Recursivecalls

visit sequence= {2, 8, 0, 9, 1, 3, 4, 5, 6, 7}

Example

0 Adjacency List

source

Visited Table (T/F)

RDFS( 2 ) RDFS(8)

RDFS(9) RDFS(1)

RDFS(3) RDFS(5) RDFS(6) no recursive call

Recursivecalls

visit sequence= {2, 8, 0, 9, 1, 3, 4, 5, 6, 7}

Example

0 Adjacency List

source

Visited Table (T/F)

RDFS(9) RDFS(1)

RDFS(3) RDFS(5) no recursive call

Recursivecalls

visit sequence= {2, 8, 0, 9, 1, 3, 4, 5, 6, 7}

Example

0 Adjacency List

source

Visited Table (T/F)

RDFS(9) RDFS(1)

Recursivecalls

visit sequence= {2, 8, 0, 9, 1, 3, 4, 5, 6, 7}

Example

0 Adjacency List

source

Visited Table (T/F)

RDFS(9) RDFS(1) no recursive call

Recursivecalls

visit sequence= {2, 8, 0, 9, 1, 3, 4, 5, 6, 7}

Example

0 Adjacency List

source

Visited Table (T/F)

RDFS( 2 ) RDFS(8)

Recursivecalls

visit sequence= {2, 8, 0, 9, 1, 3, 4, 5, 6, 7}

Example

0 Adjacency List

source

Visited Table (T/F)

RDFS( 2 ) RDFS(8) no recursive callRecursive

visit sequence= {2, 8, 0, 9, 1, 3, 4, 5, 6, 7}

Example

0 Adjacency List

source

Visited Table (T/F)

RDFS( 2 ) no recursive call Recursivecalls

visit sequence= {2, 8, 0, 9, 1, 3, 4, 5, 6, 7}

Recover a path

0 Adjacency List

source

Visited Table (T/F)

Predvisit sequence= {2, 8, 0, 9, 1, 3, 4, 5, 6, 7}

Try some examples.Path(0) ->Path(6) ->Path(7) ->

DFS Tree The edges that we traverse during DFS (or the

edges that we backtrack along) form a tree. We usually call the rooted version (rooted at the source) the DFS tree.

Minimum Spanning Trees

Problem: Laying Telephone Wire

Central office

Wiring: Naïve Approach

Central office

Expensive!

Wiring: Better Approach

Central office

Minimize the total length of wire connecting the customers

Minimum Spanning Tree (MST)(see Weiss, Section 24.2.2)

it is a tree (i.e., it is acyclic)

it covers all the vertices V contains |V| - 1 edges

the total cost associated with tree edges is the minimum among all possible spanning trees

not necessarily unique

A minimum spanning tree is a subgraph of an undirected weighted graph G, such that

Spanning Tree

• Definition– A spanning tree of a graph G is a tree (acyclic) that

connects all the vertices of G once• i.e. the tree “spans” every vertex in G

– A Minimum Spanning Tree (MST) is a spanning tree on a weighted graph that has the minimum total weight

w T w u vu v T

( ) ( , ),

such that w(T) is minimum

Where might this be useful? Can also be used to approximate someNP-Complete problems

Sample MST

• Which links to make this a MST?6

Optimal substructure: A subtree of the MST must in turn be a MST of thenodes that it spans. Will use this idea more in dynamic programming.

Kruskal’s MST Algorithm

• Idea: – Go through the list of edges and make a forest

that is a MST– At each vertex, sort the edges– Edges with smallest weights examined and

possibly added to MST before edges with higher weights

– Edges added must be “safe edges” that do not ruin the tree property.

Kruskal’s Example

• A={ }, Make each element its own set. {a} {b} {c} {d} {e} {f} {g} {h}• Sort edges.• Look at smallest edge first: {c} and {f} not in same set, add it to A, union together.• Now get {a} {b} {c f} {d} {e} {g} {h}

Kruskal ExampleKeep going, checking next smallest edge. Had: {a} {b} {c f} {d} {e} {g} {h}{e} <> {h}, add edge.

Now get {a} {b} {c f} {d} {e h} {g}

Kruskal ExampleKeep going, checking next smallest edge.Had: {a} {b} {c f} {d} {e h} {g} {a} <> {c f}, add edge.

Now get {b} {a c f} {d} {e h} {g}

Kruskal’s ExampleKeep going, checking next smallest edge. Had {b} {a c f} {d} {e h} {g}{b} <> {a c f}, add edge.

Now get {a b c f} {d} {e h} {g}

Kruskal’s ExampleKeep going, checking next smallest edge. Had {a b c f} {d} {e h} {g}{a b c f} = {a b c f}, dont add it!

Kruskal’s ExampleKeep going, checking next smallest edge. Had {a b c f} {d} {e h} {g}{a b c f} = {e h}, add it.

Now get {a b c f e h} {d}{g}

Kruskal’s ExampleKeep going, checking next smallest edge. Had {a b c f e h} {d}{g}{d} <> {a b c e f h}, add it.

Now get {a b c d e f h} {g}

Kruskal’s ExampleKeep going, check next two smallest edges. Had {a b c d e f h} {g}{a b c d e f h} = {a b c d e f h}, don’t add it.

Kruskal’s Example

Do add the last one:Had {a b c d e f h} {g}

Kruskal’s Algorithm

Kruskal(G,w) ; Graph G, with weights w A {} ; Our MST starts empty

for each vertex v V G [ ] do Make-Set(v) ; Make each vertex a set

Sort edges of E by increasing weight for each edge ( , )u v E in order

; Find-Set returns a representative (first vertex) in the set do if Find-Set(u) Find-Set(v) then A A{( , )}u v

Union(u,v) ; Combines two trees return A

Prim’s ExampleExample: Graph given earlier. Q={ (e,0) (a, ) (b, ) (c, ) (d, ) (f, ) (g, ) (h, ) }

Extract min, vertex e. Update neighbor if in Q and weight < key.

Prim’s Example

Q={ (a, ) (b,14) (c, ) (d, ) (f, ) (g, ) (h,3) } Extract min, vertex h. Update neighbor if in Q and weight < key

Prim’s Algorithm

Q={ (a, ) (b,10) (c, ) (d, ) (f,8) (g, ) } Extract min, vertex f. Update neighbor if in Q and weight < key

Prim’s Algorithm

Q={ (a, ) (b,10) (c, 2) (d, ) (g,15) } Extract min, vertex c. Update neighbor if in Q and weight < key

Prim’s Algorithm

Q={ (a,4) (b,5) (d,9) (g,15) } Extract min, vertex a. No keys are smaller than edges from a (4>2 on edge ac, 6>5 on edge ab) so nothing done. Q={ (b,5) (d,9) (g,15) } Extract min, vertex b. Same case, no keys are smaller than edges, so nothing is done. Same for extracting d and g, and we are done.

Prim’s AlgorithmGet spanning tree by connecting nodes with their parents:

Prim’s MST Algorithm• Will find a MST but may differ from Prim’s if multiple MST’s are possible

MST-Prim(G,w,r) ; Graph G, weights w, root r Q V[G]

for each vertex u Q do key[u] ; infinite “distance”

key[r] 0 P[r] NIL while Q<>NIL do u Extract-Min(Q) ; remove closest node ; Update children of u so they have a parent and a min key val ; the key is the weight between node and parent for each vAdj[u] do if vQ & w(u,v)<key[v] then P[v] u key[v] w(u,v)

Shortest Path AlgorithmsGoal: Find the shortest path between vertices in a weighted graph. We denote the shortest path between vertex u and v as (u,v). This is a very practical problem - the weights may represent the shortest distance, shortest time, shortest cost, etc. There are several forms of this problem: 1. Single-source shortest path. Find the shortest distance from a source vertex s to every other vertex in

the graph. 2. Single-destination shortest path. Find a shortest path to a given destination vertex t from every vertex

v. This is just the reverse of single-source. 3. Single-pair shortest path. Find a shortest path between a pair of vertices. No algorithm is known for

this problem that runs asymptotically faster than the best single-source algorithms. 4. All-pairs shortest path. Find the shortest path between every vertex in the graph. Note that BFS computes the shortest path on an unweighted graph.

Shortest Path?Example: What is the shortest path from g to b?

Shortest Path We will keep track of the parents of a vertex in P(v) then we can output either a shortest path using

parents or a shortest path tree. A shortest path tree is a subset of the graph with the shortest path for every vertex from a source (root).

This is not unique. We won’t use negative weights – requires a different algorithm, since negative cycles can be travelled

infinitely to make the weight cost lower and lower.

Ex: Can travel the a,b,c loop over and over again, each time reducing weight cost by one!

One way to address this problem is to shift edge values up to remove negative values

Relaxation ExampleExample: If we have that the distance from f to b is 15 (going through the direct edge), the process of relaxation updates the d[f] to be 7 going through c.

15 to 7

Relax(u,v,w) if d[v]>d[u]+w(u,v) then d[v] d[u]+w(u,v) ; decrement distance P[v] u ; indicate parent node

Dijkstra Example (0)

Dijkstra Example 1

Extract min, vertex f. S={f}. Update shorter paths.

Initialize nodes to , parent to nil. S={}, Q={(a,) (b,) (c, ) (d, ) (e, ) (f,0) (g, ) (h, )}

INF,NIL

INF,NIL INF,NIL

INF,NIL

Dijkstra Example 2

Extract min, vertex c. S={fc}. Update shorter paths.

INF,NIL

2,f 4,f

Q={(a,) (b,15) (c, 2) (d, 4) (e, ) (g, 15) (h, )} INF,NIL

Dijkstra Example 3

Extract min, vertex d. S={fcd}. Update shorter paths (None)

INF,NIL

2,f 3,c

6,cQ={(a,6) (b,7) (d, 3) (e, ) (g, 15) (h, )}

Dijkstra Example 4

INF,NIL

2,f 3,c

Extract min, vertex a. S={fcda}. Update shorter paths (None)Extract min, vertex b. S={fcdab}. Update shorter paths.

Dijkstra Example 5

Extract min, vertex h. S={fcdabh}. Update shorter paths

INF,NIL

2,f 3,c

Q={ (e, ) (g, 15) (h, 13)}

Dijkstra Example 6

Extract min, vertex g and h – nothing to update, done!

2,f 3,c

Q={ ( (e, 16) (g, 15) }

Dijkstra Example 7• Can follow parent “pointers” to get the path

2,f 3,c

Dijkstra’s AlgorithmDijkstra(G,w,s) ; Graph G, weights w, source s for each vertex vG, set d[v] and P[v] NIL d[s] 0 S{} QAll Vertices in G with associated d while Q not empty do uExtract-Min(Q) SS{u} for each vertex v Adj[u] do

if d[v]>d[u]+w(u,v) then d[v]d[u]+w(u,v) ; decrement distance

P[v] u ; indicate parent node

Dijkstra’s algorithm• S = {1} • for i = 2 to n do D[i] = C[1,i] if there is an edge from 1 to i,

infinity otherwise• for i = 1 to n-1

{ choose a vertex w in V-S such that D[w] is min add w to S (where S is the set of visited nodes) for each vertex v in V-S do D[v] = min(D[v], D[w]+c[w,v])}

Where |V| = n

all leaves are on the bottom level. all internal nodes (except perhaps the root node) have at least...

new node

revised node

current node

new root node

old parent node

leaf nodes

wellbalanced tree

median q

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