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Assume that a Binary Search Tree is constructed by repeatedly inserting exact values in to the tree. Argue that the number of nodes examined in searching for a value in the tree is one + the number of nodes determined when the value was first inserted in to the tree
Let us take an element x to insert in a binary search tree. So for inserting the
element x first at level 0,then level 1 and assume upto level (d-1). While determine at (d-1) level, x may have less or more in comparison to element at (d-1). The we enter x either left or right. In both cases no of determined code will be d. Now Assume we require to search for x, this time again traverses the similar path as we traverse. Whereas inserting the element, we stop at (d-1) the level but for searching we determine node at dth level also i.e. the node having x. Thus number of node determine while inserting are d while incase of searching it is d+1 i.e. one more than whereas inserting, as the result.
A function REPAT is specified below. Function REPAT(c in Char, i in Int, s in mString) return in mString pre 1 ≤ i ≤ the length of s. post The returned value is a string identic
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how to get plus asterisk pattern
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