<p>You are given the <code>root</code> of a binary tree containing digits from <code>0</code> to <code>9</code> only.</p>

<p>Each root-to-leaf path in the tree represents a number.</p>

<ul>
	<li>For example, the root-to-leaf path <code>1 -&gt; 2 -&gt; 3</code> represents the number <code>123</code>.</li>
</ul>

<p>Return <em>the total sum of all root-to-leaf numbers</em>. Test cases are generated so that the answer will fit in a <strong>32-bit</strong> integer.</p>

<p>A <strong>leaf</strong> node is a node with no children.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/num1tree.jpg" style="width: 212px; height: 182px;" />
<pre>
<strong>Input:</strong> root = [1,2,3]
<strong>Output:</strong> 25
<strong>Explanation:</strong>
The root-to-leaf path <code>1-&gt;2</code> represents the number <code>12</code>.
The root-to-leaf path <code>1-&gt;3</code> represents the number <code>13</code>.
Therefore, sum = 12 + 13 = <code>25</code>.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/num2tree.jpg" style="width: 292px; height: 302px;" />
<pre>
<strong>Input:</strong> root = [4,9,0,5,1]
<strong>Output:</strong> 1026
<strong>Explanation:</strong>
The root-to-leaf path <code>4-&gt;9-&gt;5</code> represents the number 495.
The root-to-leaf path <code>4-&gt;9-&gt;1</code> represents the number 491.
The root-to-leaf path <code>4-&gt;0</code> represents the number 40.
Therefore, sum = 495 + 491 + 40 = <code>1026</code>.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 1000]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 9</code></li>
	<li>The depth of the tree will not exceed <code>10</code>.</li>
</ul>

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
    int dfs_withval(TreeNode* root,int val) {
        if (root == nullptr)
            return 0;
        val = val*10 + root->val;//up to down
        if (!root->left && !root->right)
            return val;//leaf node
        int left = dfs_withval(root->left,val);
        int right = dfs_withval(root->right,val);
        
        return left+right;//without val as the val is calculated at left & right childs;
    }

    int sumNumbers(TreeNode* root) { 
        return dfs_withval(root, 0); 
    }
};
11100