Leetcode: 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.

Understand the problem:
The problem asks for the sum of all root-to-leaf numbers. The problem itself is not hard to understand. Use DFS is the natural way. The only thing to note is the overflow problem, so we may use a long long to store the intermediate results.

Solution:

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    long pathSum = 0;
    public int sumNumbers(TreeNode root) {

        sumNumbersHelper(root, 0);
        
        return (int)pathSum;
    }
    
    private void sumNumbersHelper(TreeNode root, long curSum) {
        if (root == null) {
            return;
        }
        
        curSum = curSum * 10 + root.val;
        
        if (root.left == null && root.right == null) {
            pathSum += curSum;
        }

        sumNumbersHelper(root.left, curSum);
        sumNumbersHelper(root.right, curSum);
    }
}

Update on 10/8/14:

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int sumNumbers(TreeNode root) {

        return sumNumbersHelper(root, 0);
    }
    
    private int sumNumbersHelper(TreeNode root, int preSum) {
        if (root == null) {
            return 0;
        }
        
        int curSum = root.val + preSum * 10;
        
        if (root.left == null && root.right == null) {
            return curSum;
        }
        
        return sumNumbersHelper(root.left, curSum) + sumNumbersHelper(root.right, curSum);
    }
}