Data Str & Algorithm HW Solutions 32

Data Str & Algorithm HW Solutions 32 - T By the...

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5 Binary Trees 5.1 Consider a non-full binary tree. By de f nition, this tree must have some inter- nal node X with only one non-empty child. If we modify the tree to remove X , replacing it with its child, the modi f ed tree will have a higher fraction of non-empty nodes since one non-empty node and one empty node have been removed. 5.2 Use as the base case the tree of one leaf node. The number of degree-2 nodes is 0, and the number of leaves is 1. Thus, the theorem holds. For the induction hypothesis, assume the theorem is true for any tree with n 1 nodes. For the induction step, consider a tree T with n nodes. Remove from the tree any leaf node, and call the resulting tree
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Unformatted text preview: T . By the induction hypothesis, T has one more leaf node than it has nodes of degree 2. Now, restore the leaf node that was removed to form T . There are two possible cases. (1) If this leaf node is the only child of its parent in T , then the number of nodes of degree 2 has not changed, nor has the number of leaf nodes. Thus, the theorem holds. (2) If this leaf node is the child of a node in T with degree 2, then that node has degree 1 in T . Thus, by restoring the leaf node we are adding one new leaf node and one new node of degree 2. Thus, the theorem holds. By mathematical induction, the theorem is correct. 32...
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