Finding the cheapest path down a binary tree?

Question

I'm struggling to find an algorithm for the following problem:

Given a binary tree of integers, the cost of a branch (a.k.a. a branch that starts from the root and reaches the leaf node) is given by the sum of his values. Write a function that returns a list of the cheapest branch.

Can anyone recommend me the easiest way to complete this exercise?

Answer 1

It can easily be done recursively . The function prints all the root to leaf paths along with the cheapest branch. I have used Arraylist to add all the nodes from root to leaf. Whenever a leaf node is reached i just check if the maxSum so far is smaller than then the current root to leaf path and update it.

class Node {

    public int info;
    public Node left;
    public Node right;

    public Node(int info) {
        this(info, null, null);
    }

    public Node(int info, Node left, Node right) {
        this.info = info;
        this.left = left;
        this.right = right;
    }

}

public class RootToLeaf {

    private static int maxSum = Integer.MAX_VALUE;
    private static ArrayList<Integer> finalList = new ArrayList<>();

    public static void main(String[] args) {

        Node root = new Node(8);
        root.left = new Node(4);
        root.left.left = new Node(3);
        root.left.right = new Node(1);
        root.right = new Node(5);
        root.right.right = new Node(11);
        ArrayList<Integer> list = new ArrayList<Integer>();
        path(root, list,0);
        System.out.println("Cheapest list is - " + finalList.toString() +  " and minimum sum is " + maxSum);

    }

    private static void path(Node root, ArrayList<Integer> list,int s) {

        if(root==null) {
            return;
        } else {
            list.add(root.info);
            s = s+root.info;
        }

        if ((root.left == null && root.right == null)) {
            System.out.println(list);
            if(maxSum>s) {
                maxSum = s;
                finalList = new ArrayList<>(list);
            }
            return;
        }

        path(root.left, new ArrayList<Integer>(list),s);
        path(root.right, new ArrayList<Integer>(list),s);

    }

}

The out put is as follows :

[8, 4, 3]
[8, 4, 1]
[8, 5, 11]
Cheapest list is - [8, 4, 1] and minimum sum is 13

Answer 2

As a hint, work from the leaves of the tree upward. The cost of a leaf is just the value inside the leaf. Otherwise, the cost of a the best path starting at a node is given by the cost of that node plus the cost of the cheapest path taken from there. Can you implement this recursively?

Hope this helps!