Time complexity of binary tree traversal( pre order)

time complexity of postorder traversal of binary tree
preorder traversal
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        PreorderTraversal(root)----T(n)
        {
          //Base case
          If(root equals null)-------------------O(1)
             return errormsg; ------------------O(1)
         else
             print root;---------------------------------O(1)
            //Recursive case
            PreorderTraversal(root.left);--T(n/2)
           PreorderTraversal(root.right)-T(n/2);
         }

Time complexity calculated based on back substitution method

Base case assumed as 1

        T(1)=O(1)
        T(n) = O(1)+T(n/2)
        T(n/2)=O(1)+T(n/4)
        T(n/4)=O(1)+T(n/8)
        T(n)=1+T(n/2)
               =1+1+T(n/4)
               =2+T(n/4)
               =2+1+T(n/8)
               =3+T(n/8)
        Pattern follws like
              = K+T(n/2^k).     I.e. n/2^k =1
                                             n = 2^k
                                             K= logn
              =log n + 1
              =log n

In my understanding time complexity of binary tree(pre order traversal) is log n most of them said time complexity of binary tree(pre order) is O(n)

I don't know which one is correct??

Kindly guide me for above issue??

The time complexity for pre-order traversal is O(n) because you visit each node only once.

Tree Traversals (Inorder, Preorder and Postorder), In case of binary search trees (BST), Inorder traversal gives nodes in non- decreasing order. To get nodes Preorder traversal is also used to get prefix expression on of an expression tree. Please see Time Complexity: O(n) Preorder traversal of binary tree is 1 2 4 5 3 Inorder traversal of binary tree is 4 2 5 1 3 Postorder traversal of binary tree is 4 5 2 3 1. One more example: Time Complexity: O(n) Let us see different corner cases. Complexity function T(n) — for all problem where tree traversal is involved — can be defined as:

The reason you get T(n)=log(n) is that your equation for complexity is incorrect. It should be T(n)=O(1)+2T(n/2) (because you are applying the same algorithm for the left and the right child). And this will be the case only when the tree is balanced, if the tree is linear then you would have T(n)=O(1)+T(n-1) which will give you clearly a linear time. Also the first case will give you a linear time. By applying your method you would get T(n) = O(2^k * T(1)) where k = log(n) so T(n)=O(n).

To make it easy to compute time complexity just imagine what the algorithm does, here the preorder will visit all the nodes, and each node only once, so it is intuitive that is O(n) time complexity

Complexities of binary tree traversals, 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),� For all of these traversals - whether done recursively or iteratively - you’ll have to visit every node in the binary tree. That means that you’ll get a runtime complexity of [math]O(n)[/math] - where n is the number of nodes in the binary tree.

I had the same question today. Your equation is slightly wrong. That's why you're getting log(n) complexity.

T(n) = T(n/2) + T(n/2) = 2T(n/2) //equation 1
T(1) = O(1) //base case
T(n/2) = T(n/4) + T(n/4) = 2T(n/4) //equation 2

On solving them, we get,

T(n) = 2T(n/2) 
T(n) = 2*2(T(n/4)) = 4(T(n/4))
T(n) = 2^k * T(n/2^k) //equation 3

On solving for the value of k, we get,

n/2^k = 1
n = 2^k
log n = k

On using the value of k in equation 3, we get,

T(n) = 2^logn * T(1) // 2^logn is n.
T(n) = O(n) * O(1)
T(n) = O(n)

And that is how you get O(n) time complexity.

What are the time complexities of preorder, inorder, postorder, and , What are the time complexities of preorder, inorder, postorder, and level order binary tree traversals with both recursive and iterative algorithms? In-order traversal is very commonly used on binary search trees because it returns values from the underlying set in order, according to the comparator that set up the binary search tree. Post-order traversal while deleting or freeing nodes and values can delete or free an entire binary tree. Thereby the node is freed after freeing its children.

[PDF] CS 261: Data Structures Binary Tree Traversals Binary Search Trees, visit nodes in a tree? • Most common tree traversal orders: –Pre-order. –In- order. –Post-order. 2 Complexity of Binary Tree Traversals requires O(n) time . 8� Example Input: Inorder= [D, B, E, A, F, C] Preorder= [A, B, D, E, C, F] Output: Pre-order traversal of the tree formed by the given preorder and inorder A B D E C F In-order traversal of the tree formed by the given preorder and inorder D B E A F C Post-order traversal of the tree formed by the given preorder and inorder D E B F C A

Calculation of Inorder Traversal Complexity, In order to analyse the time complexity of a tree traversal you have to think in the terms of number of nodes visited. If a tree has n nodes, then� To prove that the time complexity is O (n) O(n) O (n), the biggest problem lies in finding the time complexity of finding the predecessor nodes of all the nodes in the binary tree. Intuitively, the complexity is O ( n log ⁡ n ) O(n\log n) O ( n lo g n ) , because to find the predecessor node for a single node related to the height of the tree.

Tree traversal, Such traversals are classified by the order in which the nodes are visited. The following algorithms are described for a binary tree, but they may be generalized � Pre-order Traversal. For now, if we already have an idea about how the in-order traversal works in the binary tree, we can flexibly adapt to another type of traversal, which is the pre-order traversal, in the pre-order traversal, we explore the tree as follow: Visit the node; Recursively explore the left subtree; Recursively explore the right