> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# Smallest Subtree with all the Deepest Nodes

> Tested Python solution for LeetCode 865 with 16 pytest cases. Generate a practice environment with lcpy.

LeetCode 865, [Medium](/catalog/medium). Topics: [Hash Table](/catalog/topics/hash-table), [Tree](/catalog/topics/tree), [Depth-First Search](/catalog/topics/depth-first-search), [Breadth-First Search](/catalog/topics/breadth-first-search), [Binary Tree](/catalog/topics/binary-tree), Binary Lifting, Lowest Common Ancestor, DP on Trees. [View on LeetCode](https://leetcode.com/problems/smallest-subtree-with-all-the-deepest-nodes/description/).

Generate this problem as a practice environment: tested reference solution, 16 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 865   # by problem number
lcpy gen -s smallest_subtree_with_all_the_deepest_nodes   # by problem name
```

## Problem

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 a tree consisting of that node, plus the set of all descendants of that node.

### Examples

![Example 1](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/07/01/sketch1.png)

```
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.
```

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

```
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].
* 0 \<= Node.val \<= 500
* The values of the nodes in the tree are unique.

**Note:** This question is the same as 1123: [https://leetcode.com/problems/lowest-common-ancestor-of-deepest-leaves/](https://leetcode.com/problems/lowest-common-ancestor-of-deepest-leaves/)

## Solution

Reference implementation from [solution.py on GitHub](https://github.com/wislertt/leetcode-py/blob/main/leetcode/smallest_subtree_with_all_the_deepest_nodes/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/smallest_subtree_with_all_the_deepest_nodes/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
from leetcode_py import TreeNode


class Solution:
    # Time: O(n) single DFS over all nodes
    # Space: O(h) recursion stack, h = tree height
    def subtree_with_all_deepest(self, root: TreeNode[int] | None) -> TreeNode[int] | None:
        if root is None:
            return None

        def dfs(node: TreeNode[int]) -> tuple[int, TreeNode[int]]:
            left = dfs(node.left) if node.left else (0, node)
            right = dfs(node.right) if node.right else (0, node)
            if left[0] > right[0]:
                return left[0] + 1, left[1]
            if left[0] < right[0]:
                return right[0] + 1, right[1]
            return left[0] + 1, node

        return dfs(root)[1]
```

## Complexity

| Time | Space |
| - | - |
| O(n) single DFS over all nodes | O(h) recursion stack, h = tree height |

## Tags


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