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57 lines
1.5 KiB
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57 lines
1.5 KiB
Plaintext
---
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title: '235. Lowest Common Ancestor of a Binary Search Tree'
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description: Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST
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sidebar:
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label: 'Lowest Common Ancestor of a Binary Search Tree'
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badge: 'Medium'
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---
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<Badge variant="accent">Binary Search Tree</Badge>
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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 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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## Solution
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```py
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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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```
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