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Assume a complete binary tree T with n nodes where each node has an item (value). Label the nodes of the complete binary tree T from top to bottom & from left to right 0, 1, ..., n-1. Relate with T the array A where the ith entry of A is the item in the node labeled i of T, i = 0, 1, ..., n-1. Table illustrates the array representation of a Binary tree of Figure
Given the index i of a node, we can efficiently & easily compute the index of its parent and left & right children:
Index of Parent: (i - 1)/2, Index of Left Child: 2i + 1, Index of Right Child: 2i + 2.
Node #
Item
Left child
Right child
0
A
1
2
B
3
4
C
-1
D
5
6
E
7
8
G
H
I
J
9
?
Table: Array Representation of a Binary Tree
First column illustrates index of node, second column contain the item stored into the node & third & fourth columns mention the positions of left & right children
(-1 shows that there is no child to that specific node.)
* Initialise d & pi* for each vertex v within V( g ) g.d[v] := infinity g.pi[v] := nil g.d[s] := 0; * Set S to empty * S := { 0 } Q := V(g) * While (V-S)
The above 3 cases are also considered conversely while the parent of Z is to the right of its own parent. All the different kind of cases can be illustrated through an instance. Le
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null(nil) = true // nil refer for empty tree null(fork(e, T, T'))= false // e : element , T and T are two sub tree leaf(fork(e, nil, nil)) = true leaf(
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