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vb.net print barcode zebra COUNTING BINARY TREES in Java
COUNTING BINARY TREES Scanning DataMatrix In Java Using Barcode Control SDK for Java Control to generate, create, read, scan barcode image in Java applications. Make Data Matrix In Java Using Barcode generator for Java Control to generate, create Data Matrix image in Java applications. Figure 11.3 A binary tree
Scan DataMatrix In Java Using Barcode scanner for Java Control to read, scan read, scan image in Java applications. Bar Code Encoder In Java Using Barcode maker for Java Control to generate, create bar code image in Java applications. EXAMPLE 11.2 All the Binary Trees of Size 3
Scan Barcode In Java Using Barcode recognizer for Java Control to read, scan read, scan image in Java applications. ECC200 Creation In C# Using Barcode generation for Visual Studio .NET Control to generate, create ECC200 image in .NET applications. There are five different binary trees of size n = 3, as shown in Figure 11.4.
DataMatrix Creation In Visual Studio .NET Using Barcode generator for ASP.NET Control to generate, create DataMatrix image in ASP.NET applications. Painting Data Matrix In VS .NET Using Barcode encoder for VS .NET Control to generate, create Data Matrix image in Visual Studio .NET applications. Figure 11.4 The five different binary trees of size 3
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UPCA Generation In Java Using Barcode encoder for Java Control to generate, create UPC Code image in Java applications. Encoding Bar Code In Java Using Barcode generator for Java Control to generate, create barcode image in Java applications. EXAMPLE 11.3 All the Binary Trees of Size 4
Printing USD8 In Java Using Barcode creation for Java Control to generate, create Code11 image in Java applications. UPCA Recognizer In None Using Barcode recognizer for Software Control to read, scan read, scan image in Software applications. There are 14 different binary trees of size n = 4, as shown in Figure 11.5.
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Code 39 Full ASCII Encoder In None Using Barcode maker for Software Control to generate, create Code 3 of 9 image in Software applications. UCC.EAN  128 Drawer In Java Using Barcode creation for BIRT reports Control to generate, create UCC  12 image in BIRT reports applications. Ten have height 3, and the other four have height 2.
Drawing Code39 In VS .NET Using Barcode creator for Reporting Service Control to generate, create Code39 image in Reporting Service applications. ECC200 Creator In None Using Barcode drawer for Online Control to generate, create Data Matrix image in Online applications. EXAMPLE 11.4 All the Binary Trees of Size 5
To find all the binary trees of size 5, apply the recursive definition for binary trees. If t is a binary tree of size 5, then it must consist of a root node together with two subtrees the sum of whose sizes equals 4. There are four possibilities: The left subtree contains either 4, 3, 2, 1, or 0 nodes. First count all the binary trees of size 5 whose left subtree has size 4. From Example 11.3, we see that there are 14 different possibilities for that left subtree. But for each of those 14 choices, there are no other options because the right subtree must be empty. Therefore, there are 14 different binary trees of size 5 whose left subtree has size 4. Next, count all the binary trees of size 5 whose left subtree has size 3. From Example 11.2, we see that there are five different possibilities for that left subtree. But for each of those five choices, there are no BINARY TREES
[CHAP. 11
other options because the right subtree must be a singleton. Therefore, there are five different binary trees of size 5 whose left subtree has size 3. Next, count all the binary trees of size 5 whose left subtree has size 2. There are only two different possibilities for that left subtree. But for each of those two choices, we have the same two different possibilities for the right subtree because it also must have size 2. Therefore, there are 2 2 = 4 different binary trees of size 5 whose left subtree has size 2. By similar reasoning, we find that there are five different binary trees of size 5 whose left subtree has size 1, and there are 14 different binary trees of size 5 whose left subtree has size 0. Therefore, the total number of different binary trees of size 5 is 14 + 5 + 4 + 5 + 14 = 42. FULL BINARY TREES A binary tree is said to be full if all its leaves are at the same level and every interior node has two children. EXAMPLE 11.5 The Full Binary Tree of Height 3 The tree shown in Figure 11.6 is the full binary tree of height 3. Note that it has 15 nodes: 7 interior nodes and 8 leaves. Figure 11.6 A full binary tree of height 3
Theorem 11.1 The full binary tree of height h has l = 2h leaves and m = 2h 1 internal nodes. Proof: The full binary tree of height h = 0 is a single leaf node, so it has n = 1 node, which is a leaf. Therefore, since 2h 1 = 20 1 = 1 1 = 0, and 2h = 20 = 1, the formulas are correct for the case where h = 0. More generally, let h > 0 and assume (the inductive hypothesis) that the formulas are true for all full binary trees of height less than h. Then consider a full binary tree of height h. Each of its subtrees has height h 1, so we apply the formulas to them: lL = lR = 2h 1 and mL = mR = 2h 1 1. (These are the number of leaves in the left subtree, the number of leaves in the right subtree, the number of internal nodes in the left subtree, and the number of internal nodes in the right subtree, respectively.) Then l = lL + lR = 2h 1 + 2h 1 = 2 2h 1 = 2h and m = mL + mR + 1 = (2h 1 1) + (2h 1 1) + 1 = 2 2h 1 1 = 2I 1 Therefore, by the (Second) Principle of Mathematical Induction, the formulas must be true for full binary trees of any height h 0. By simply adding the formulas for m and l, we obtain the first corollary. Corollary 11.1 The full binary tree of height h has a total of n = 2h+1 1 nodes. By solving the formula n = 2h+1 1 for h, we obtain this corollary: Corollary 11.2 The full binary tree with n nodes has height h = lg(n+1) 1. Note that the formula in Corollary 11.2 is correct even in the special case where n = 0: The empty binary tree has height h = lg(n+1) 1 = lg(0+1) 1 = lg(1) 1 = 0 1 = 1. CHAP. 11]

