Given the <code>root</code> of a binary tree, return <em>the sum of values of its deepest leaves</em>.
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/07/31/1483_ex1.png" style="width: 273px; height: 265px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,4,5,null,6,7,null,null,null,null,8]
<strong>Output:</strong> 15
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = [6,7,8,2,7,1,3,9,null,1,4,null,null,null,5]
<strong>Output:</strong> 19
</pre>

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 10<sup>4</sup>]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 100</code></li>
</ul>

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def deepestLeavesSum(self, root: Optional[TreeNode]) -> int:
        ans = 0
        mx = -1
        def dfs(node, depth):
            nonlocal mx, ans
            if not node:
                return
            if depth == mx:
                ans += node.val
            elif depth > mx:
                mx, ans = depth, node.val
            dfs(node.left, depth+1)
            dfs(node.right, depth+1)
            

        dfs(root, 0)
        return ans
        
19200