mirror of
https://github.com/prdlk/leetcode.git
synced 2026-09-17 07:26:27 +00:00
71 lines
1.8 KiB
Python
71 lines
1.8 KiB
Python
|
"""
|
||
|
|
235. Lowest Common Ancestor of a Binary Search Tree
|
||
|
|
Difficulty: Medium
|
||
|
|
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, 10^5].
|
||
|
|
|
||
|
|
• -10^9 <= Node.val <= 10^9
|
||
|
|
|
||
|
|
• All Node.val are unique.
|
||
|
|
|
||
|
|
• p != q
|
||
|
|
|
||
|
|
• p and q will exist in the BST.
|
||
|
|
"""
|
||
|
|
|
||
|
|
# 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":
|
||
|
|
cur = root
|
||
|
|
while cur:
|
||
|
|
if p.val < cur.val and q.val < cur.val:
|
||
|
|
cur = cur.left
|
||
|
|
elif p.val > cur.val and q.val > cur.val:
|
||
|
|
cur = cur.right
|
||
|
|
else:
|
||
|
|
return cur
|