Leetcode & CTCI ---Day 5

Sum Root to Leaf Numbers

 Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path 1->2->3 which represents the number 123.

Find the total sum of all root-to-leaf numbers.

For example,

    1
   / \
  2   3

 

The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.

Return the sum = 12 + 13 = 25.

My initial solution used an arrayList to hold all root to leaf numbers, and then finally add them all. 

 

public class Solution {
    ArrayList<Integer> result = new ArrayList<Integer>();
    public int sumNumbers(TreeNode root) {
        if (root == null)
            return 0;
        sumNumbersHelper(root, 0);
        int total = 0;
        for (int a : result){
            total += a;
        }
        return total;
    }
    
    public void sumNumbersHelper(TreeNode root, int sum){
        if (root == null)
            return;
        sum = sum * 10 + root.val;
        if (root.left == null && root.right == null){
            result.add(sum);
            return;
        }
        else {
            if (root.left != null)
                sumNumbersHelper(root.left, sum);
            if (root.right != null)
                sumNumbersHelper(root.right, sum);
        }
    }
    
}

 

 

 

But later I found this much cleaner solution, do not need an arrayList. Instead, just sum them directly.

public class Solution {
    int result = 0;
    public int sumNumbers(TreeNode root) {
        if (root == null)
            return 0;
        sumNumbersHelper(root, 0);
        return result;
    }
    
    public void sumNumbersHelper(TreeNode root, int sum){
        if (root == null)
            return;
        sum = sum * 10 + root.val;
        if (root.left == null && root.right == null){
            result += sum;
            return;
        }
        else {
            if (root.left != null)
                sumNumbersHelper(root.left, sum);
            if (root.right != null)
                sumNumbersHelper(root.right, sum);
        }
    }
    
}