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Implement exponentiation and link tree neighbors
This question evaluates numerical algorithm implementation and in-place tree traversal skills, specifically fast exponentiation with careful edge-case and precision handling and iterative linking of next pointers in a perfect binary tree.
Solve two independent tasks.
Task A — Fast exponentiation: Implement fastExponent(x, n) that returns x raised to the integer power n, where x is a double and n is a 32-bit signed integer. Requirements: O(log |n|) time via exponentiation by squaring; O(1) extra space with an iterative approach; correctly handle negative exponents, n == Integer.MIN_VALUE, and edge cases such as x == 0 or x == ±1. Return a double; a relative error up to 1e-10 is acceptable.
Task B — Link neighbors in a perfect binary tree: Given the root of a perfect binary tree whose nodes have fields (val, left, right, next), set each node’s next pointer to its immediate right neighbor on the same level; if none, set it to null. Constraints: O(n) time, O(1) extra space, and do not use recursion. Afterward, each level should be traversable using next pointers starting from its leftmost node.
Fast Exponentiation
Return x raised to integer n using exponentiation by squaring.
Examples
Example 1
Input: (2.0, 10)
Output: 1024.0
Explanation: Positive exponent.
Example 2
Input: (2.0, -2)
Output: 0.25
Explanation: Negative exponent.
Link Neighbors in a Perfect Binary Tree
For a perfect tree represented level-order, return each node's next value or None.
Examples
Example 1
Input: ([1, 2, 3, 4, 5, 6, 7],)
Output: [None, 3, None, 5, 6, 7, None]
Explanation: Three levels.
Example 2
Input: ([1],)
Output: [None]
Explanation: Single root.