In a binary max heap containing n numbers, the smallest element can be found in time (GATE CS 2006)
(A) 0(n)
(B) O(logn)
(C) 0(loglogn)
(D) 0(1)


Answer: (A)

Explanation: In a max heap, the smallest element is always present at a leaf node. So we need to check for all leaf nodes for the minimum value. Worst case complexity will be O(n)

         12
        /  \
      /      \
    8         7
   / \        / \
 /     \    /     \
2      3   4       5

Quiz of this Question