Maximum Sum Path

July 22, 2014

Sometimes here at Programming Praxis we publish exercises taken from homework in computer science classes, sometimes we publish exercises that are based on or related to problems at Project Euler, and sometimes we publish exercises that based on common technical interview questions. Today we hit the trifecta: a common interview question that is frequently assigned as homework in computer science classes and that also appears at Project Euler (twice!):

Given a binary tree with integers stored in its nodes, find the maximum sum of any path from root to leaf.

For instance, in the binary tree represented by the triangle shown below, the maximum sum path goes from the 3 at the root, to its child 7, to its grandchild 4, to the leaf node 9, a sum of 23. Since there is no other path with a greater sum, the maximum sum is 23:

   3
  7 4
 2 4 6
8 5 9 3

Your task is to write a program to compute the maximum sum path through a binary tree; also write a variant of the program that returns the path. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

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4 Responses to “Maximum Sum Path”

  1. Paul said

    In Python.

    import StringIO
    DATA = """75
    95 64
    17 47 82
    18 35 87 10
    20 04 82 47 65
    19 01 23 75 03 34
    88 02 77 73 07 63 67
    99 65 04 28 06 16 70 92
    41 41 26 56 83 40 80 70 33
    41 48 72 33 47 32 37 16 94 29
    53 71 44 65 25 43 91 52 97 51 14
    70 11 33 28 77 73 17 78 39 68 17 57
    91 71 52 38 17 14 91 43 58 50 27 29 48
    63 66 04 68 89 53 67 30 73 16 69 87 40 31
    04 62 98 27 23 09 70 98 73 93 38 53 60 04 23"""
    
    def bottom_up(data):
        while len(data) > 1:
            bottom = data.pop()
            data[-1][:] = [t + max(bottom[i:i+2]) for i, t in enumerate(data[-1])]
        return data[0][0]
        
    def bottom_up_path(data):
        data[-1][:] = [(t, [t]) for t in data[-1]]
        while len(data) > 1:
            bottom = data.pop()
            for i, t in enumerate(data[-1]):
                m = max(bottom[i:i+2])
                data[-1][i] = (t + m[0], [t] + m[1])
        return data[0][0]
        
    data = [[int(x) for x in line.split()] for line in StringIO.StringIO(DATA)]
    print bottom_up_path(data)
    # (1074, [75, 64, 82, 87, 82, 75, 73, 28, 83, 32, 91, 78, 58, 73, 93])
    
  2. matthew said

    In C++: read data from stdin to an stl::vector, check it’s triangular and compute what the row size is. Loop over vector filling in maximum subtree sizes, then we can retrieve the path fairly easily by scanning the data again from the top (no need to keep track of paths as we fill in table):

    #include <iostream>
    #include <vector>
    #include <math.h>
    #include <assert.h>
    using namespace std;
    // 2A = N(N+1) => N^2+N-2A = 0
    // a = 1, b = 1, c = -2A
    // N = (sqrt(1+8A)-1)/2
    int main()
    {
       vector<int> a;
       {
          int n;
          while(cin >> n) {
             a.push_back(n);
          }
       }
       int A = a.size();;
       int N = (sqrt(1+8*A)-1)/2; // IEEE754 sqrt needed!
       assert(A == N*(N+1)/2);    // Check the input is sane
       {
          int q = A-N-1; // The table position we are filling in.
          for (int n = N-1; n > 0; n--) { // The row widths
             for (int i = n; i > 0; i--,q--) { // Go across the row
                a[q] += max(a[q+n],a[q+n+1]);
             }
          }
       }
       {
          int p = 0;    // current index in path
          for (int n = 1; n < N; n++) {
             int t = a[p]; // current value
             p += n+(a[p+n+1] > a[p+n]);
             cout << t-a[p] << " ";
          }
          cout << a[p] << "\n";
       }
    }
    
    $ ./maxsumpath <<EOF
    75
    95 64
    17 47 82
    18 35 87 10
    20 04 82 47 65
    19 01 23 75 03 34
    88 02 77 73 07 63 67
    99 65 04 28 06 16 70 92
    41 41 26 56 83 40 80 70 33
    41 48 72 33 47 32 37 16 94 29
    53 71 44 65 25 43 91 52 97 51 14
    70 11 33 28 77 73 17 78 39 68 17 57
    91 71 52 38 17 14 91 43 58 50 27 29 48
    63 66 04 68 89 53 67 30 73 16 69 87 40 31
    04 62 98 27 23 09 70 98 73 93 38 53 60 04 23
    EOF
    75 64 82 87 82 75 73 28 83 32 91 78 58 73 93
    
  3. My Scala solution can be found here.

  4. svenningsson said

    Haskell solution:

    module MaximumSumPath where
    
    data Tree a = Node a (Tree a) (Tree a) | Leaf
    
    maxSum = maxSum' 0
    maxSum' p Leaf = p
    maxSum' p (Node i l r) = max (maxSum' (p+i) l) (maxSum' (p+i) r)
    
    data Choice = L | R deriving Show
    
    maxSumPath = (s,reverse path)
      where (s,path) = maxSumPath' (0,[])
    maxSumPath' p Leaf = p
    maxSumPath' (s,p) (Node i l r) = maxPath (maxSumPath' (s+i,L:p) l)
                                             (maxSumPath' (s+i,R:p) r)
      where maxPath p1@(s1,_) p2@(s2,_) | s1 > s2 = p1
                                        | otherwise = p2

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