Created
July 31, 2017 19:58
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Get maximum independent branch sum
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public class MaxSumInTree{ | |
int left(int i) { | |
return (2 * i + 1); | |
} | |
int right(int i) { | |
return (2 * i + 2); | |
} | |
long findMax(int n, String tree) { | |
String[] data = tree.split("\\s+"); | |
Arrays.stream(data).forEach(System.out::println); | |
return findMaxRecursive(data, 0, data[0]); | |
} | |
long findMaxRecursive(String[] data, int i, String parent) { | |
if (parent.equals("#")) { | |
return 0; | |
} | |
if (left(i) >= data.length) { | |
return Long.parseLong(parent); | |
} | |
long a = findMaxRecursive(data, left(i), data[i]); | |
long b = findMaxRecursive(data, right(i), data[i]); | |
if (i == 0) { | |
return a + b; | |
} else { | |
return Math.max(a, b); | |
} | |
} | |
public static void main(String[] args) { | |
String data = "3 4 5 1 3 # 1"; | |
System.out.println(" The max independent sum in binary tree is " + new MaxSumInTree().findMax(6, data)); | |
} | |
} |
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