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Recursion-1 > sumDigits
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Given a non-negative int n, return the sum of its digits recursively (no loops). Note that mod (%) by 10 yields the rightmost digit (126 % 10 is 6), while divide (/) by 10 removes the rightmost digit (126 / 10 is 12).


sumDigits(126) → 9
sumDigits(49) → 13
sumDigits(12) → 3

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public int sumDigits(int n) { }

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