Chapter 21 Binary Heap
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Chapter 21Binary Heap
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Objective
To learn Basic properties of the binary heap Insert and deletemin operations Linear time heap-construction algorithm Heapsort
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Topics
Review and analyze operations on Heap data structure
Review definitions Tree, binary treedepthheight full binary treecomplete binary treeHeap property
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Binary trees A tree is an acyclic, connected, undirected
graph.
Only one path exists between a pair of nodes
A leaf of a tree is a node with no children.
Binary tree - a tree where each node has 0,1 or 2 children
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Depth and height of a tree Depth of a node is:
Depth of the root of a tree is 0.The depth of its parent +1
Depth of a tree is maximum depth of its leaves.
Height of a node is:Height of a leaf of a tree is 0.The maximum height of its children +1
Height of a tree is the height of the root.
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Depth and height of a binary tree
0
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2
2
1 0
0 0
Depth
The depth of the tree = height of the tree = 2
Height
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A complete binary tree
A full binary tree (also called complete) is a binary tree such thatAll internal nodes have 2 childrenAll leaves have depth d
A complete binary tree (also called essentially complete) is a binary tree such thatAll internal nodes have 2 children except the last
internal node which may have only 1 child.All leaves have depth d or d -1Nodes with depth d are as far to the left as
possible.
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Full binary tree Complete binary tree
n = 2h+1 -1 2h n 2h+1 -1
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The height of a complete binary tree
The number of nodes n of a complete binary tree satisfies: 2h n (2h+1-1)
Taking the log base 2 we get:
h lg n and lg(n+1) h+1 or
lg(n+1)-1 h lg n
Since h is integer h = lg(n + 1)-1= lg n
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The height of a complete binary tree
2h n (2h+1-1)
Full binary treeall leaves same depth
Complete binarytree with 1 node at depth h
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Heap Definition
A heap is a complete binary tree that satisfies the heap property.
Minimum Heap Property: The value stored at each node is less than or equal
to the values stored at its children.
OR Maximum Heap Property: greater
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Implementation of Heap For simplicity we assume the complete
binary tree is an array, and the root is stored at index 1.
For any element in array position i, its left child is at position 2i, and it’s right child is at 2i+1, its parent is at i/2 .
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Last node in a heap
The last node of a heap is the rightmost internal node of on the last level
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last node
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Heap viewed as Binary tree implemented as an array.
3 2
8 7
18 14 9
29 6
1
23
4 5 6 7
8 9 10
1
1 3 2 8 7 29 6
1 2 3 4 5 6 7 8 9
18 14 9
10
last
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insert(v )
Item inserted as last item in the heapHeap property may
be violated
Do percolate to restore heap property
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65
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insertion node
2
65
79 1
z
z
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Percolate up After the insertion of a new key k, the heap-order property
may be violated Algorithm percolate up restores the heap-order property by
swapping k along an upward path from the insertion node percolate up terminates when the key k reaches the root or a
node whose parent has a key smaller than or equal to k Since a heap has height O(log n), upheap runs in O(log n)
time
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15
79 6z
1
25
79 6z
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deleteMin()
Save root object O(1)
Remove last element and store in root O(1)
siftDown(1)
30 20
80
1
2 3
4
10
30 20
1
2 3
80
30 80
1
2 3
20
10
last
18
4 3
8 17
12 14 19
6 13
1
23
4 5 6 7
8 9 10
9siftDown(1)New value at root.
Satisfy the Heap property.
Right Child is smallerExchange root and right child
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4 9
8 17
12 14 19
6 13
1
23
4 5 6 7
8 9 10
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ParentLeft Child is smallerExchange parent and left child
20
4 6
8 17
12 14 19
9 13
1
23
4 5 6 7
8 9 10
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The worst case run time to do siftDown(index) isO(h(index)) where h(index) is the height of node indexWhen index=root=1, O(lg n)
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Build heap---linear time//Build-heap
1. for i (last /2)downto 1
2. do siftDown( i )
Why we can start siftDown at last/2 ?
because we :need to siftDown only parents the rest of the nodes are leaves and leaves satisfy the heap property
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5 8 12 9 7 10 21
1 2 3 4 5 6 7 8 9
6 14 4
10
8 12
9 7
6 14 4
10 21
1
23
4 5 6 7
8 9 10
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siftDown makes it a min heap
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8 12
9 4
6 14 7
10 21
1
23
4 5 6 7
8 9 10
5
i = 4
siftDown makes this into heap
this is a heap
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8 12
6 4
9 14 7
10 21
1
23
4 5 6 7
8 9 10
5
These are heaps
i = 3
siftDown makes heap
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1
siftDown
i = 2
10
12 21
3
6 7
5
8
6 4
9 14 7
2
4 5
8 9 10
4
6 8
9 14 7
2
4 5
8 9 10
4
6 7
9 14 8
2
4 5
8 9 10
26
4 10
6 7
9 14 8
12 21
1
23
4 5 6 7
8 9 10
5i = 1
10
6 7
9 14 8
12 21
1
23
4 5 6 7
8 9 10
4
5
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Running time
n/2 nodes For each node, at most O(log(n)) The running time is O(nlgn)
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Heap-Sort
Using a heap-based priority queue, we can sort a sequence of n elements in O(n log n) time
The resulting algorithm is called heap-sort
Heap-sort is much faster than quadratic sorting algorithms, such as insertion-sort
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97
5953
412658 31
16 3621 97 53 59 26 41 58
1 2 3 4 50
31 16 21 36
6 7 8 9 10
59 53 58 26 41 36
1 2 3 4 50
31 16 21 97
6 7 8 9 10
59
5853
412636 31
1621 97
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HeapSort
//build heap
for i (last /2)downto 1
do siftDown( a, i, last) //shiftDown from i to last
for j last-1 downto 1
swap(a[0], a[j]) //deleteMax
do siftDown(a, 0, j) // shiftDown from 0 to j
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Common errors (from book, page 786) For heapsort data begins in position 0, so
the children of the I are in positions 2i+1 and 2i+2.
STL priority queue is a max heap, not a min heap.
More errors are defined in the book.
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In class exercises
21.1 Describe the structure and ordering properties of the
binary heap 21.2
In a binary heap, for an item in position I where are the parent, left child, and right child located?
21.3 Show the result of inserting 10, 12, 1, 14, 6, 5, 8, 15,
3, 9, 7, 4, 11, 13, and 2, one at a time, in an initially empty heap. Then show the result of using the linear-time buildHep algorithm instead
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