<p>Given the <code>root</code> of a binary tree, return the leftmost value in the last row of the tree.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/12/14/tree1.jpg" style="width: 302px; height: 182px;" />
<pre>
<strong>Input:</strong> root = [2,1,3]
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/12/14/tree2.jpg" style="width: 432px; height: 421px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,4,null,5,6,null,null,7]
<strong>Output:</strong> 7
</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>-2<sup>31</sup> &lt;= Node.val &lt;= 2<sup>31</sup> - 1</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 helper(self, node, depth):
        if node.left is None and node.right is None:
            return node.val, depth
        elif node.left is not None and node.right is None:
            return self.helper(node.left, depth + 1)
        elif node.left is None and node.right is not None:
            return self.helper(node.right, depth + 1)
        else:
            leftVal, leftDepth = self.helper(node.left, depth + 1)
            rightVal, rightDepth = self.helper(node.right, depth + 1)
            if leftDepth < rightDepth:    return rightVal, rightDepth
            else:   return leftVal, leftDepth

    def findBottomLeftValue(self, root: Optional[TreeNode]) -> int:
        if root.left is None and root.right is None:
            return root.val
        else:
            val, depth = self.helper(root, 1)
            return val
19800