Python solution
First use BFS to find the deepest nodes, then find the lowest common ancestor (LCA) of these nodes. \n\nTime complexity: O(n), space complexity: O(n).\n\nclass 
First use BFS to find the deepest nodes, then find the lowest common ancestor (LCA) of these nodes. \n\nTime complexity: O(n), space complexity: O(n).\n```\nclass Solution:\n    def subtreeWithAllDeepest(self, root):\n        """\n        :type root: TreeNode\n        :rtype: TreeNode\n        """\n        if not root:\n            return\n\t\t\t\n\t\t# BFS to find the deepest nodes\n        q = collections.deque([(root,0)])\n        maxdepth = -float(\'inf\')\n        res = []\n        dic = {} # keep track of the parent\n        while q:\n            u, depth = q.popleft()\n            if depth > maxdepth:\n                maxdepth = depth\n                res = [u]\n            elif depth == maxdepth:\n                res.append(u)\n            if u.left:\n                q.append((u.left, depth+1))\n                dic[u.left] = u\n            if u.right:\n                q.append((u.right, depth+1))\n                dic[u.right] = u\n\t\t\t\t\n\t\t# find the LCA of nodes in res\n        if len(res) == 1:\n            return res[0]\n        else:\n            while True:\n                parent = []\n                for u in res:\n                    if not parent:\n                        parent.append(dic[u])\n                    else:\n                        if dic[u] != parent[0]:\n                            parent.append(dic[u])\n                if len(parent) == 1:\n                    return parent[0]\n                else:\n                    res = parent\n```
[]