human
Solve in Python:
Given the root of a binary tree, the depth of each node is the shortest distance to the root.
Return the smallest subtree such that it contains all the deepest nodes in the original tree.
A node is called the deepest if it has the largest depth possible among any node in the entire tree.
The subtree of a node is tree consisting of that node, plus the set of all descendants of that node.
 
Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4]
Output: [2,7,4]
Explanation: We return the node with value 2, colored in yellow in the diagram.
The nodes coloured in blue are the deepest nodes of the tree.
Notice that nodes 5, 3 and 2 contain the deepest nodes in the tree but node 2 is the smallest subtree among them, so we return it.

Example 2:
Input: root = [1]
Output: [1]
Explanation: The root is the deepest node in the tree.

Example 3:
Input: root = [0,1,3,null,2]
Output: [2]
Explanation: The deepest node in the tree is 2, the valid subtrees are the subtrees of nodes 2, 1 and 0 but the subtree of node 2 is the smallest.

 
Constraints:

The number of nodes in the tree will be in the range [1, 500].
The values of the nodes in the tree are unique.
assistant
#Here is the Python code you requested.

# 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 subtreeWithAllDeepest(self, root: TreeNode) -> TreeNode:
        
        
        def getDepth(root):
            if not root:
                return 0
            left=getDepth(root.left)
            right=getDepth(root.right)
            return 1+max(left,right)
        
        def getSubtrees(root):
            if root:
                
                leftDepth=getDepth(root.left)
                rightDepth=getDepth(root.right)
                if leftDepth==rightDepth:
                    return root
                elif leftDepth>rightDepth:
                    return getSubtrees(root.left)
                else:
                    return getSubtrees(root.right)
                
        return getSubtrees(root)