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:
        # Can we have more than 2 leaves? 
        # traverse tree as BFS
        # for each layer - if we found a new leaf - replace the last calculated sum with a new one 
        deepest_sum = 0

        queue = deque([root])

        while queue:
            num_nodes = len(queue)
            deepest_sum = 0
            for _ in range(num_nodes):
                n = queue.popleft()

                if not n.left and not n.right:
                    deepest_sum += n.val

                for x in [n.left, n.right]:
                    if x:
                        queue.append(x)

        return deepest_sum            
        
19400