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.

Note: A leaf is a node with no children.

Example:

Input: [1,2,3]
    1
   / \
  2   3
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.

Example 2:

Input: [4,9,0,5,1]
    4
   / \
  9   0
 / \
5   1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.
 /**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int sumNumbers(TreeNode root) {
        return checkNodeSum(root, 0);
    }

    public int checkNodeSum(TreeNode node, int sum) {
        if (node == null) return 0;
        if (node.left == null && node.right == null) return sum * 10 + node.val;
        return checkNodeSum(node.left, sum * 10 + node.val) + checkNodeSum(node.right, sum * 10 + node.val);
    }
}


Here, we are going to solve using recursion. We have created another method which contains TreeNode and sum as arguments. Initially sum is 0.

1) If root is null, return 0.
2) If both left and right of root are null then, return sum * 10 + node value.
3)Else recursively call left tree and its sum and add it with recusively calling right tree with its sum.


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