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71 lines
1.8 KiB
Python
71 lines
1.8 KiB
Python
"""
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235. Lowest Common Ancestor of a Binary Search Tree
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Difficulty: Medium
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https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/
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──────────────────────────────────────────────────
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Given a binary search tree (BST), find the lowest common ancestor
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(LCA) node of two given nodes in the BST.
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According to the definition of LCA on Wikipedia: “The lowest
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common ancestor is defined between two nodes p and q as the lowest
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node in T that has both p and q as descendants (where we allow a node
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to be a descendant of itself).”
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Example 1:
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Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
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Output: 6
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Explanation: The LCA of nodes 2 and 8 is 6.
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Example 2:
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Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
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Output: 2
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Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a
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descendant of itself according to the LCA definition.
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Example 3:
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Input: root = [2,1], p = 2, q = 1
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Output: 2
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Constraints:
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• The number of nodes in the tree is in the range [2, 10^5].
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• -10^9 <= Node.val <= 10^9
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• All Node.val are unique.
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• p != q
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• p and q will exist in the BST.
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"""
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution:
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def lowestCommonAncestor(
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self, root: "TreeNode", p: "TreeNode", q: "TreeNode"
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) -> "TreeNode":
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cur = root
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while cur:
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if p.val < cur.val and q.val < cur.val:
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cur = cur.left
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elif p.val > cur.val and q.val > cur.val:
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cur = cur.right
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else:
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return cur
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