235. 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, 105].
• -109 <= Node.val <= 109
• All Node.val are unique.
• p != q
• p and q will exist in the BST.

---

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

'''
1. 아이디어 :
    1) while문으로 노드를 하나씩 내려가면서, p, q가 해당 노드보다 값이 작으면 노드를 왼쪽으로, 반대면 오른쪽으로 가고,
    둘다 아니면 양쪽으로 가기때문에 해당 노드를 리턴하면 된다.
2. 시간복잡도 :
    1) O(logn)
3. 자료구조 :
    1) 이진트리
'''


# 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':
        curr = root
        while curr:
            if p.val < curr.val and q.val < curr.val:
                curr = curr.left
            elif p.val > curr.val and q.val > curr.val:
                curr = curr.right
            else:
                return curr