Complexities of binary tree traversals

Question

What is the time complexity of inorder postorder and preorder traversal of binary trees in data structures   Is it O n  or O log n  or O n 2

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T n    2T n 2   c   T n 2    2T n 4    c    T n    4T n 4    2c   c   similarly               T n    8T n 8    4c  2c   c              last step     T n    nT 1    c sum of powers of 2 from 0 to h height of tree    so Complexity is O 2  h 1  -1   but h   log n    so  O 2n - 1    O n

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O n   because you traverse each node once  Or rather - the amount of work you do for each node is constant  does not depend on the rest of the nodes

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Depth first traversal of a binary tree is of order O n   Algo --  lt b gt      PreOrderTrav   -----------------T n  lt b gt      if root is null---------------O 1  lt b gt        return null-----------------O 1  lt b gt      else -------------------------O 1  lt b gt        print root -----------------O 1  lt b gt        PreOrderTrav root left -----T n 2  lt b gt        PreOrderTrav root right ----T n 2  lt b gt   If the time complexity of the algo is T n  then it can be written as T n    2 T n 2    O 1   If we apply back substitution we will get T n    O n

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O n  I would say   I am doing for a balanced tree applicable for all the trees  Assuming that you use recursion   T n    2 T n 2    1 ----------   1   T n 2  for left sub-tree and T n 2  for right sub-tree and  1  for verifying the base case   On Simplifying  1  you can prove that the traversal either inorder or preorder or post order  is of order O n

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In-order  Pre-order  and Post-order traversals are Depth-First traversals   For a Graph  the complexity of a Depth First Traversal is O n   m   where n is the number of nodes  and m is the number of edges   Since a Binary Tree is also a Graph  the same applies here  The complexity of each of these Depth-first traversals is O n m    Since the number of edges that can originate from a node is limited to 2 in the case of a Binary Tree  the maximum number of total edges in a Binary Tree is n-1  where n is the total number of nodes   The complexity then becomes O n   n-1   which is O n

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Consider a skewed binary tree with 3 nodes as 7  3  2  For any operation like for searching 2  we have to traverse 3 nodes  for deleting 2 also  we have to traverse 3 nodes and for for inserting 1 also  we have to traverse 3 nodes  So  binary tree has worst case complexity of O n

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Travesal is O n  for any order - because you are hitting each node once   Lookup is where it can be less than O n  IF the tree has some sort of organizing schema  ie binary search tree

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Introduction  Hi  I was asked this question today in class  and it is a good question   I will explain here and hopefully get my more formal answer reviewed or corrected where it is wrong      Previous Answers  The observation by  Assaf is also correct since binary tree traversal travels recursively to visit each node once   But   since it is a recursive algorithm  you often have to use more advanced methods to analyze run-time performance    When dealing with a sequential algorithm or one that uses for-loops  using summations will often be enough   So  what follows is a more detailed explanation of this analysis for those who are curious   The Recurrence  As previously stated   T n    2 T n 2    1  where T n  is the number of operations executed in your traversal algorithm  in-order  pre-order  or post-order makes no difference   Explanation of the Recurrence  There are two T n  because inorder  preorder  and postorder traversals all call themselves on the left and right child node   So  think of each recursive call as a T n    In other words    left T n 2    right T n 2    2 T n 2       The  1  comes from any other constant time operations within the function  like printing the node value  et cetera    It could honestly be a 1 or any constant number  amp  the asymptotic run-time still computes to the same value   Explanation follows     Analysis  This recurrence actually can be analyzed using big theta using the masters  theorem   So  I will apply it here   T n    2 T n 2     constant  where constant is some constant  could be 1 or any other constant    Using the Masters  Theorem   we have  T n    a T n b    f n    So  a 2  b 2  f n    constant since f n    n c   1  then it follows that c   0 since f n  is a constant   From here  we can see that a   2 and b c   2  0   1   So  a b c or 2 2 0   So  c  lt  logb a  or 0  lt  log2 2   From here we have T n    BigTheta n  logb a      BigTheta n 1    BigTheta n   If your not famliar with BigTheta n   it is  similar    please bear with me      to O n  but it is a  tighter bound  or tighter approximation of the run-time   So  BigTheta n  is both worst-case O n   and best-case BigOmega n  run-time   I hope this helps   Take care

[time-complexity] Complexities of binary tree traversals

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