Balanced Binary Search Tree - NJU
Transcript of Balanced Binary Search Tree - NJU
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Balanced Binary Search Tree
Algorithm : Design & Analysis[8]
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In the last class…
Finding max and minFinding the second largest keyAdversary argument and lower boundSelection Problem – MedianA Linear Time Selection AlgorithmAnalysis of Selection AlgorithmA Lower Bound for Finding the Median
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Balanced Binary Search Tree
Definition of red-black treeBlack heightInsertion into a red-black treeDeletion from a red-black tree
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Binary Search Tree Revisited
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Poor balancingΘ(n)
•Each node has a key, belonging to a linear ordered set•An inorder traversal produces a sorted list of the keys
•Each node has a key, belonging to a linear ordered set•An inorder traversal produces a sorted list of the keys
In a properly drawn tree, pushing forward to get the ordered list.
Good balancingΘ(logn)
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Node Group in a binTree
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5 principal subtrees5 principal subtrees
Node groupNode group
As in 2-tree, the number of external node is one more than that of internal node
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Improving the Balancing by Rotation
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Root of the group is changed.
The middle principal subtree changes parent
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Red-Black Tree: the Definition
If T is a binary tree in which each node has a color, red or black, and all external nodes are black, then T is a red-black tree if and only if:
[Color constraint] No red node has a red child[Black height constrain] The black length of all external paths from a given node u is the same (the black height of u)The root is black.
Almost-red-black tree(ARB tree)Root is red, satisfying the other constraints.
Balancing is under controlledBalancing is under controlled
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Recursive Definition of Red-Black Tree
(A red-black tree of black height h is denoted as RBh)
Definition:An external node is an RB0 tree, and the node is black.A binary tree is an ARBh (h≥1) tree if:
Its root is red, andIts left and right subtrees are each an RBh-1 tree.
A binary tree is an RBh (h≥1) tree if:Its root is black, andIts left and right subtrees are each either an RBh-1 tree or an ARBhtree.
No ARB0
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RBi and ARBi
RB0
ARB1
RB1
(1) (2)
(4)(3)
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Red-Black Tree with 6 Nodes40
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poorest balancing: height(normal) is 4
Black edge
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Black-Depth Convention
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All with the same largest black depth: 2All with the same
largest black depth: 2
ARB TreesARB Trees
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Properties of Red-Black Tree
The black height of any RBh tree or ARBh tree is well defind and is h.
Let T be an RBh tree, then:T has at least 2h-1 internal black nodes.T has at most 4h-1 internal nodes.The depth of any black node is at most twice its black depth.
Let A be an ARBh tree, then:A has at least 2h-2 internal black nodes.A has at most (4h)/2-1 internal nodes.The depth of any black node is at most twice its black depth.
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Well-Defined Black HeightThat “the black height of any RBh tree or ARBh tree is well defind”means the black length of all external paths from the root is the same. Proof: induction on hBase case: h=0, that is RB0 (there is no ARB0)In ARBh+1, its two subtrees are both RBh.Since the root is red, the black length of all external paths from the root is h, that’s the same as its two subtrees.In RBh+1:
Case 1: two subtrees are RBh’sCase 2: two subtrees are ARBh+1’sCase 3: one subtree is an RBh(black height=h), and the another is an ARBh+1(black height=h+1)
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Bound on Depth of Node in RBTree
Let T be a red-black tree with n internal nodes. Then no node has depth greater than 2lg(n+1), which means that the height of T in the usual sense is at most 2lg(n+1).
Proof:Let h be the black height of T. The number of internal nodes, n, is at least the number of internal black nodes, which is at least 2h-1, so h≤lg(n+1). The node with greatest depth is some external node. All external nodes are with black depth h. So, the depth is at most 2h.
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Influences of Insertion into an RB TreeBlack height constrain:
No violation if inserting a red node.Color constraint:
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Critical clusters(external nodes excluded), which originated by color violation, with 3 or 4 red nodes
Critical clusters(external nodes excluded), which originated by color violation, with 3 or 4 red nodes
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Repairing 4-node Critical Cluster
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Color flip:Root of the critical cluster exchanges color with its subtrees
Color flip:Root of the critical cluster exchanges color with its subtrees
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No new critical cluster occurs, inserting finished.
No new critical cluster occurs, inserting finished.
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Repairing 4-node Critical Cluster2 more insertions
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Critical cluster
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New critical cluster with 3 nodes.Color flip doesn’t work,Why?
New critical cluster with 3 nodes.Color flip doesn’t work,Why?
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Patterns of 3-Node Critical Cluster
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Shown as properly drawn
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Repairing 3-Node Critical Cluster
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RRoot of the critical cluster is changed to M, and the parentship is adjusted accordingly
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The incurred critical cluster is of pattern A
All into one pattern
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Implementing Insertion: Class
class RBtreeElement root;RBtree leftSubtree;RBtree rightSubtree;int color; /* red, black */
static class InsReturnpublic RBtree newTree;public int status /* ok, rbr, brb, rrb, brr */
class RBtreeElement root;RBtree leftSubtree;RBtree rightSubtree;int color; /* red, black */
static class InsReturnpublic RBtree newTree;public int status /* ok, rbr, brb, rrb, brr */
Color pattern
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Implementing Insertion: ProcedureRBtree rbtInsert (RBtree oldRBtree, Element newNode)
InsReturn ans = rbtIns(oldREtree, newNode);If (ans.newTree.color ≠ black)
ans.newTree.color = black;return ans.newTree;
RBtree rbtInsert (RBtree oldRBtree, Element newNode)InsReturn ans = rbtIns(oldREtree, newNode);If (ans.newTree.color ≠ black)
ans.newTree.color = black;return ans.newTree;
InsReturn rbtIns(RBtree oldRBtree, Element newNode)InsReturn ans, ansLeft, ansRight;if (oldRBtree = nil) then <Inserting simply>;else
if (newNode.key <oldRBtree.root.key)ansLeft = rbtIns (oldRBtree.leftSubtree, newNode);ans = repairLeft(oldRBtree, ansLeft);
else ansRight = rbtIns(oldRBtree.rightSubtree, newNode);ans = repairRight(oldRBtree, ansRight);
return ans the recursive function
the wrapper
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Correctness of InsertionIf the parameter oldRBtree of rbtIns is an RBh tree or an ARBh+1 tree(which is true for the recursive calls on rbtIns), then the newTree and status fields returned are one of the following combinations:
Status=ok, and newTree is an RBh or an ARBh+1 tree,Status=rbr, and newTree is an RBh, Status=brb, and newTree is an ARBh+1 tree,Status=rrb, and newTree.color=red, newTree.leftSubtree is an ARBh+1tree and newTree.rightSubtree is an RBh tree,Status=brr, and newTree.color=red, newTree.rightSubtree is an ARBh+1tree and newTree.leftSubtree is an RBh tree
For those cases with red root, the color will be changed to black, with other constraints satisfied by repairing subroutines.
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Deletion: Logical and Structralu: to be deleted logicallyu: to be deleted logically
σ: tree successor of u, to be deleted structurally, with information moved into u
σ: tree successor of u, to be deleted structurally, with information moved into u
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π
S right subtree of S, to replacing S
right subtree of S, to replacing S
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After deletion
π is parent of σπ is parent of σ
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Deletion from RBTree: Examples
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Original tree
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Deletion in a Red-Black Tree
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To be deletedTo be deleted
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π The black height of πis not well-defined !
black depth=1
black depth=2
one deletion
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Procedure of Red-Black Deletion
1. Do a standard BST search to locate the node to be logically deleted, call it u
2. If the right child of u is an external node, identify u as the node to be structurally deleted.
3. If the right child of u is an internal node, find the tree successor of u , call it σ, copy the key and information from σ to u. (color of u not changed) Identify σ as the node to be deleted structurally.
4. Carry out the structural deletion and repair any imbalance ofblack height.
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Imbalance of Black Height60
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Analysis of Black Imbalance
The imbalance occurs when:A black node is delete structurally, and Its right subtree is black (external)
The result is:An RBh-1 occupies the position of an RBh as required by its parent, coloring it as a “gray” node.
Solution:Find a red node and turn it black as locally as possible.The gray color might propagate up the tree.
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Propagation of Gray Node
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Map of the vicinity of g, the gray node
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The pattern for which propagation is needed
g-subtree gets well-defined black height, but that is less than that required by its parent
g-subtree gets well-defined black height, but that is less than that required by its parent
Gray UpIn the worst case, up to the root of the tree, and successful
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Repairing without Propagation
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Deletion Rebalance group
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Restructured
Restructuring the deletion rebalance group:•Red p: form an RB1 or ARB2 tree•Black p: form an RB2 tree
Restructuring the deletion rebalance group:•Red p: form an RB1 or ARB2 tree•Black p: form an RB2 tree
RB2
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Repairing without Propagation
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Deletion Rebalance group
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Restructured
Restructuring the deletion rebalance group:•Red p: form an RB1 or ARB2 tree•Black p: form an RB2 tree
Restructuring the deletion rebalance group:•Red p: form an RB1 or ARB2 tree•Black p: form an RB2 tree
ARB2
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Complexity of Operations on RBTree
With reasonable implementationA new node can be inserted correctly in a red-black tree with n nodes in Θ(logn) time in the worst case.Repairs for deletion do O(1) structural changes, but may do O(logn) color changes.
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