vb.net code to generate barcode Figure 11.31 An array in Java

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Figure 11.31 An array
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The natural mapping of the specified binary tree is shown in Figure 11.32.
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Figure 11.32 An array
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CHAP. 11]
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BINARY TREES
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The natural mapping of the specified binary tree is shown in Figure 11.33.
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Figure 11.33 An array
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11.8 11.9
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The level order traversal will print the numbers from the natural mapping in order. The expression tree for a*(b + c)*(d*e + f) is shown in Figure 11.34.
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Figure 11.34 A binary tree
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11.10 11.11
The prefix expression is *a*+bc+*def. The postfix expression is *abc+de*f +**. Figure 11.35 shows the expression tree for each of the prefix expressions given in Problem 5.2 on page 111.
Figure 11.35 Prefix expression trees
Figure 11.36 shows the expression tree for each of the infix expressions given in Problem 5.4 on page 111.
Figure 11.36 Infix expression trees
BINARY TREES
[CHAP. 11
Figure 11.37 shows the expression tree for each of the postfix expressions given in Problem 5.6 on page 111.
Figure 11.37 Postfix expression trees
11.14 11.15 11.16 11.17 11.18 11.19
Figure 11.38 shows the expression tree for a*(b + c)*(d*e + f) is: In a binary tree of height h = 4, 5 In a binary tree with n = 7 nodes, 2 n h 31. 6.
For a given number of nodes, the highest binary tree is a linear sequence. For a given number of nodes, the lowest binary tree is a complete binary tree. To verify the recursive definition for the Figure 11.38 An expression tree given tree, we first note that the leaves C, E, and F are binary trees because every singleton satisfies the recursive definition for binary trees because its left and right subtrees are both empty (and therefore binary trees). Next, it follows that the subtree rooted at B is a binary tree because it is a triplet (X,L,R) where X = B, L = , and R = C. Similarly, it follows that the subtree rooted at D is a binary tree because it is a triplet (X,L,R) where X = D, L = E, and R = F. Finally, it follows that the entire tree satisfies the recursive definition because it is a triplet (X,L,R) where X = A, L is the binary tree rooted at B, and L is the binary tree rooted at D. Figure 11.39 on page 227 shows all 42 different binary trees of size n = 5. There are 132 different binary trees of size 6: 1 42 + 1 14 + 2 5 + 5 2 + 14 1 + 42 1 = 132. A nonempty binary tree consists of a root X, a left subtree L, and a right subtree R. Let n be the size of the binary tree, let nL = |L| = the size of L, and nR = |R| = the size of R. Then n = 1 + nL + nR . So there are only n different possible values for the pair (nL , nR ): (0, n 1), (1, n 2), ..., (n 1,0). For example, if n = 6 (as in Problem 11.21), the only possibilities are (0,5), (1,4), (2,3), (3,2), (4,1), or (5,0). In the (0, n 1) case, L is empty and |R| = n 1; there are f(0) f(n 1) different binary trees in that case. In the (1, n 2) case, L is a singleton and |R| = n 2; there are f(1) f(n 2) different binary trees in that case. The same principle applies to each case. Therefore the total number of different binary trees of size n is f(n) = 1 f(n 1) + 1 f(n 2) + 2 f(n 3) + 5 f(n 4) + 14 f(n 5) + + f(i 1) f(n i) + + f(n 1) 1 In closed form, the formula is
11.20 11.21 11.22
f n =
n i=1
f i 1
f n i
CHAP. 11]
BINARY TREES
Figure 11.39 The 42 binary trees of size 5
These are called the Catalan numbers: n 2n n n +1 2n n ---------------n+1
1 1 2 5 14 42 132 429
f i 1
f n i
0 1 1 2 2 6 3 20 4 70 5 252 6 924 7 3432
1 2 3 4 5 6 7 8
1 11=1 1 1+1 1 = 2 1 2+1 1+2 1 = 5 1 5+1 2+2 1+5 1 = 14 1 14+1 5+2 2+5 1+14 1 = 42 1 42+1 14+2 5+5 2+14 1+42 1 = 132 1 132+1 42+2 14+5 5+14 2+42 1+132 1 = 429 Table 11.1 Catalan numbers
For a given height h > 0, the binary tree with the most nodes is the full binary tree. Corollary 11.1 on page 202 states that that number is n 2h+1 1. Therefore, in any binary tree of height h, the number n of nodes must satisfy n 2h+1 1. The binary tree with the fewest nodes for a given height h is the one in which every internal node has only one child; that linear tree has n = h + 1 nodes because every node except the single leaf has exactly one child. Therefore, in any binary tree of height h, the number n of nodes must satisfy n h + 1. The second pair of inequalities follows from the first by solving for h. Let T be any binary tree of height h and size n. Let T1 be the smallest complete binary tree that contains T. Let h1 be the height of T1 and let n1 be its size. Then h = h1 and n n1. Then by Corollary 11.1 on page 202, n n1 2h+1 1. The required inequalities follow from this result.
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