235. Lowest Common Ancestor of a Binary Search Tree
Medium LeetCodeGiven 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
q 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, 105]
- -109 <= Node.val <= 109
- All Node.val are unique
- p != q
- p and q will exist in the BST
How to solve the problem
Code
- Recursion
Python
# 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 p.val < root.val and q.val < root.val: # Because this is a BST, all the nodes smaller then current node are in left subtree.
return self.lowestCommonAncestor(root.left, p, q)
if p.val > root.val and q.val > root.val:
return self.lowestCommonAncestor(root.right, p, q)
return root
Complexity
- Time complexity: O(n), n == number of nodes
- Space complexity: O(n)
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