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vb.net code to generate barcode Figure 11.31 An array in Java
Figure 11.31 An array EAN13 Supplement 5 Reader In Java Using Barcode Control SDK for Java Control to generate, create, read, scan barcode image in Java applications. Encoding EAN13 In Java Using Barcode maker for Java Control to generate, create GTIN  13 image in Java applications. The natural mapping of the specified binary tree is shown in Figure 11.32.
EAN / UCC  13 Scanner In Java Using Barcode decoder for Java Control to read, scan read, scan image in Java applications. Barcode Drawer In Java Using Barcode maker for Java Control to generate, create barcode image in Java applications. Figure 11.32 An array
Barcode Decoder In Java Using Barcode scanner for Java Control to read, scan read, scan image in Java applications. Print EAN 13 In Visual C# Using Barcode printer for .NET framework Control to generate, create European Article Number 13 image in .NET framework applications. CHAP. 11] Printing EAN 13 In .NET Framework Using Barcode maker for ASP.NET Control to generate, create EAN / UCC  13 image in ASP.NET applications. Create EAN13 Supplement 5 In .NET Using Barcode creation for .NET Control to generate, create EAN / UCC  13 image in VS .NET applications. BINARY TREES
Paint EAN13 In VB.NET Using Barcode encoder for VS .NET Control to generate, create European Article Number 13 image in Visual Studio .NET applications. Encoding Barcode In Java Using Barcode creator for Java Control to generate, create bar code image in Java applications. The natural mapping of the specified binary tree is shown in Figure 11.33.
Barcode Encoder In Java Using Barcode creation for Java Control to generate, create bar code image in Java applications. EAN13 Drawer In Java Using Barcode creation for Java Control to generate, create GTIN  13 image in Java applications. Figure 11.33 An array
Paint Code 2/5 In Java Using Barcode generation for Java Control to generate, create 2/5 Industrial image in Java applications. Encoding Bar Code In .NET Using Barcode printer for Reporting Service Control to generate, create barcode image in Reporting Service applications. 11.8 11.9 Bar Code Maker In Visual Studio .NET Using Barcode generator for .NET Control to generate, create barcode image in .NET applications. Scanning Bar Code In Visual Basic .NET Using Barcode scanner for .NET framework Control to read, scan read, scan image in VS .NET applications. 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. Scan Bar Code In None Using Barcode reader for Software Control to read, scan read, scan image in Software applications. GTIN  13 Generation In Java Using Barcode drawer for Android Control to generate, create UPC  13 image in Android applications. Figure 11.34 A binary tree
Recognize EAN 13 In None Using Barcode decoder for Software Control to read, scan read, scan image in Software applications. Barcode Maker In Visual Basic .NET Using Barcode drawer for .NET framework Control to generate, create bar code image in VS .NET applications. 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.

