vb.net code to generate barcode COUNTING BINARY TREES EXAMPLE 11.2 All the Binary Trees of Size 3 in Java

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COUNTING BINARY TREES EXAMPLE 11.2 All the Binary Trees of Size 3
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Figure 11.3 A binary tree
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There are five different binary trees of size n = 3, as shown in Figure 11.4.
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Figure 11.4 The five different binary trees of size 3
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Four have height 2, and the other one has height 1.
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EXAMPLE 11.3 All the Binary Trees of Size 4
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There are 14 different binary trees of size n = 4, as shown in Figure 11.5.
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Figure 11.5 The 14 different binary trees of size 4
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Ten have height 3, and the other four have height 2.
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EXAMPLE 11.4 All the Binary Trees of Size 5
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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]
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