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. Topics: Hash Table, Tree, Depth-First Search, Breadth-First Search, Binary Tree, Binary Lifting, Lowest Common Ancestor, DP on Trees. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 16 parametrized pytest cases, and a playground notebook:

lcpy gen -n 865   # by problem number
lcpy gen -s smallest_subtree_with_all_the_deepest_nodes   # by problem name

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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.

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Examples

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.

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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/

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

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]

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Complexity

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

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