Lowest Common Ancestor of a BST

Problem

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/

Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Example 1:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
Output: 6
Explanation: The LCA of nodes 2 and 8 is 6.

Example 2:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
Output: 2
Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

Example 3:

Input: root = [2,1], p = 2, q = 1
Output: 2

Constraints:

• The number of nodes in the tree is in the range [2, 10:sup:`5`].

• -10:sup:`9`<= Node.val <= 10:sup:`9`

• All Node.val are unique.

• p != q

• p and q will exist in the BST.

Pattern

Tree, Depth-First Search, Binary Search Tree, Binary Tree

Approaches

Recursive

Code

from __future__ import annotations


class TreeNode:
    """Node in a binary tree."""

    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

    @classmethod
    def from_list(cls, vals: list[int | None]) -> TreeNode | None:
        if not vals:
            return None
        root = TreeNode(vals[0])
        queue = [root]
        i = 1
        while i < len(vals):
            node = queue.pop(0)
            if i < len(vals) and vals[i] is not None:
                node.left = TreeNode(vals[i])
                queue.append(node.left)
            i += 1
            if i < len(vals) and vals[i] is not None:
                node.right = TreeNode(vals[i])
                queue.append(node.right)
            i += 1
        return root


def lowestCommonAncestor(
    root: TreeNode | None,
    p: TreeNode,
    q: TreeNode,
) -> TreeNode | None:
    if root is None:
        return None
    elif max(p.val, q.val) < root.val:
        return lowestCommonAncestor(root.left, p, q)
    elif min(p.val, q.val) > root.val:
        return lowestCommonAncestor(root.right, p, q)
    else:
        return root

Test

>>> from lowest_common_ancestor_of_a_binary_search_tree__recursive import lowestCommonAncestor, TreeNode
>>> root = TreeNode.from_list([6, 2, 8, 0, 4, 7, 9, None, None, 3, 5])
>>> lowestCommonAncestor(root, TreeNode(2), TreeNode(8)).val
6
>>> lowestCommonAncestor(root, TreeNode(2), TreeNode(4)).val
2

class lowest_common_ancestor_of_a_binary_search_tree__recursive.TreeNode(val=0, left=None, right=None)

Bases: object

Node in a binary tree.

classmethod from_list(vals: list[int | None]) → TreeNode | None

lowest_common_ancestor_of_a_binary_search_tree__recursive.lowestCommonAncestor(root: TreeNode | None, p: TreeNode, q: TreeNode) → TreeNode | None