'what would be the run time of this function using master's theorem

    IN-TREE(p, k) 
    if p == NIL 
        return FALSE 
    else if p.key == KEY 
        return TRUE 
    else 
        return IN-TREE(p.left, k) or IN-TREE(p.right, k)

p points to a node in a complete linked binary tree and has each node p has a key called p.key, a pointer to its left subtree p.left, and a pointer to its right subtree p.right. An empty subtree is written as NIL.

What would be the run time of this function using master's theorem?

algorithmruntimebig-opseudocodemaster-theorem


Solution 1:[1]

It is clearly evident that in the worst-case scenario, you are going to look over all the nodes in the tree to match the node's key with KEY.

Therefore time complexity of this solution is O(n), where n is the number of nodes in the Tree.

Solution 2:[2]

A complete tree of depth d has 2^d-1 nodes. So in the worst case (when the key is not found),

T(d) = C + T(d-1) + T(d-1).

If we rewrite in terms of the number of nodes,

T(N) = C + 2 T((N-1)/2).

As the independent term is constant, the solution is linear in N.

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 umang-malhotra
Solution 2 Yves Daoust

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