LeetCode235. Lowest Common Ancestor of a Binary Search Tree

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本文探讨了在二叉搜索树中寻找两个指定节点的最低公共祖先（LCA）的问题，提供了详细的解决方案及代码实现。通过递归算法，利用二叉搜索树的特性，高效地找到目标节点的公共父节点。

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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]

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         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.

思路：一般二叉树的题用递归很好做。这个题首先是Binary Search Tree，就是说左子树所有的值都小于根节点的值，右子树的值都大于根节点的值。所以找共同父节点时，只需要从上向下，找第一个值处于这两个子节点值之间就可以。

# 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, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        if root.val>max(p.val, q.val):
            return self.lowestCommonAncestor(root.left, p, q)
        elif root.val<min(p.val, q.val):
            return self.lowestCommonAncestor(root.right, p, q)
        return root