leetcode 235 Lowest Common Ancestor of a Binary Search Tree 二叉搜索树的最近公共祖先 python 最简解法（迭代、递归）

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'''

Given a binary search tree (BST), find the lowest common ancestor (LCA) 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).”

Given binary search tree:  root = [6,2,8,0,4,7,9,null,null,3,5]




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.


Note:

All of the nodes' values will be unique.
p and q are different and both values 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':

        # # Approach one   递归求解
        # if p.val < root.val > q.val:
        #     return self.lowestCommonAncestor(root.left, p, q)
        # if p.val > root.val < q.val:
        #     return self.lowestCommonAncestor(root.right, p, q)
        # return root


        # Approach two   迭代求解
        while True:
            if p.val < root.val > q.val:
                root = root.left
            elif p.val > root.val < q.val:
                root = root.right
            else:
                return root

所有Leetcode题目不定期汇总在 Github, 欢迎大家批评指正，讨论交流。