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Unformatted text preview: HW5F07 CS336 1. Define recursively functions height, number of nodes, number of leaves and number of internal (interior) nodes for extended binary trees. Solution: For the height function h :: T N , we have h. h ( d, , ) = 0 h. 1 h ( d,l,r ) = 1 + max ( h ( l ) ,h ( r )) For the number of nodes # N :: T N , we have # N. 0 # N ( d, , ) = 1 # N. 1 # N ( d,l,r ) = 1 + # N ( l ) + # N ( r ) For the number of leaves # L :: T N , we have # L. 0 # L ( d, , ) = 1 # L. 1 # L ( d,l,r ) = # L ( l ) + # L ( r ) For the number of internal nodes # I :: T N , we have # I. 0 # I ( d, , ) = 0 # I. 1 # I ( d,l,r ) = 1 + # I ( l ) + # I ( r ) 2. Show that the minimum number of internal nodes for an extended binary tree of height h is h . Solution: For the base case, we have h = 0 and t = ( d, , ), and # I ( d, , ) < # I. > = 0 < h = 0 > = h For the inductive step, the I.H. is the following: for trees of height k h with the minimum number of internal nodes, we have # I = k . We need to show that for tree of height h + 1 with the minimum number of internal nodes, we have # I = h + 1. Let t = ( d,l,r ) be a binary tree of height h +1 with the minimum number of internal nodes. W.L.O.G. let l be of height h with the minimum number of internal nodes and then r must be . (Otherwise we can subtract nodes 1 from the subtree without changing the height of t and that would be a contradiction.) Then we have # I ( t ) < t = ( d,l,r ) > = # I ( d,l, ) < # I. 1 > = 1 + # I ( l ) + # I ( ) < I.H. , # I. > = 1 + h + 0 < arith > = h + 1 Hence by P.M.I., Q.E.D....
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 Spring '08
 Myers

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