Given a rooted binary tree, return the lowest common ancestor of its deepest leaves.
Recall that:
The node of a binary tree is a leaf if and only if it has no children
The depth of the root of the tree is 0, and if the depth of a node is d, the depth of each of its children is d+1.
The lowest common ancestor of a set S of nodes is the node A with the largest depth such that every node in S is in the subtree with root A.
Example 1:
Input: root = [1,2,3]
Output: [1,2,3]
Explanation: 
The deepest leaves are the nodes with values 2 and 3.
The lowest common ancestor of these leaves is the node with value 1.
The answer returned is a TreeNode object (not an array) with serialization "[1,2,3]".
Example 2:
Input: root = [1,2,3,4]
Output: [4]
Example 3:
Input: root = [1,2,3,4,5]
Output: [2,4,5]
Constraints:
The given tree will have between 1 and 1000 nodes.
Each node of the tree will have a distinct value between 1 and 1000.
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#   1
# /   \\
# 2    3

# [1, 2,null, null,3, null, null]

class Solution:
    def lcaDeepestLeaves(self, root: TreeNode) -> TreeNode:
      return self.helper(root)[0]
        
    def helper(self, root):
        if not root:
            return None, 0
        n1, l1 = self.helper(root.left)
        n2, l2 = self.helper(root.right)

        if l1 > l2:
            return n1, l1+1
        elif l1 < l2:
            return n2, l2+1
        else:
            return root, l1+1

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#   1
# /   \\
# 2    3

# [1, 2,null, null,3, null, null]

class Solution:
    def lcaDeepestLeaves(self, root: TreeNode) -> TreeNode:
      return self.helper(root)[0]
        
    def helper(self, root):
        if not root:
            return None, 0
        n1, l1 = self.helper(root.left)
        n2, l2 = self.helper(root.right)

        if l1 > l2:
            return n1, l1+1
        elif l1 < l2:
            return n2, l2+1
        else:
            return root, l1+1
