235. Lowest Common Ancestor of a Binary Search Tree

Problem

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).”

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

Example 1:

case1

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:

case2

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, 10⁵].
-10⁹ <= Node.val <= 10⁹
All Node.val are unique.
p != q
p and q will exist in the BST.

Test Cases

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':

solution_test.py

import pytest

import os
import sys
sys.path.append(os.path.dirname(os.path.dirname(__file__)))
from _utils.binary_tree import bst_find, build_tree
from solution import Solution

null = None


@pytest.mark.parametrize('root, p, q, expected', [
    ([6,2,8,0,4,7,9,null,null,3,5], 2, 8, 6),
    ([6,2,8,0,4,7,9,null,null,3,5], 2, 4, 2),
])
class Test:
    def test_solution(self, root, p, q, expected):
        sol = Solution()
        root = build_tree(root)
        p = bst_find(root, p)
        q = bst_find(root, q)
        assert sol.lowestCommonAncestor(root, p, q).val == expected

Thoughts

不妨设 p < q，显然在 BST 中，p 和 q 的 LCA 满足 p <= lca(p, q) <= q。更高的祖先要么同时小于 p 和 q，要么同时大于它俩。

从根节点出发，如果节点的值刚好介于 p 和 q 之间，就是 LCA。否则如果比 p 和 q 都大，就进入左子树；反之则进入右子树。

Code

solution.py

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        if q.val < p.val:
            p, q = q, p

        while root and not p.val <= root.val <= q.val:
            root = root.left if q.val < root.val else root.right

        return root

本文除了题目描述（Problem）部分之外，其他内容由 Calf 原创，采用 CC BY-NC-SA 4.0 许可协议，转载请注明出处。