235. Lowest Common Ancestor of a Binary Search Tree

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本文介绍了一种寻找二叉搜索树中两节点的最低公共祖先(LCA)的有效算法。利用二叉搜索树特性，通过比较节点值大小确定LCA位置，避免了遍历整棵树，提升了效率。

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 v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5

For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

这个题目首先要看清是二查搜索树，我当时忘记这一点了所以走了很多弯路，

这道题因为二查搜索树的缘故简单很多，：

1. 如果root的值在pq之间，则root就是lca

2.如果pq 的值都小于root，那么pq就在root的左子树上，自然他们的lca也在root的左子树上

3.如果pq的值都大于root，那么pq就在root的右子树上，自然他们的lca也在root的右子树上

所以最后的程序是：

class Solution(object):
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        if p.val == q.val:
        	return p
        elif p.val >q.val:
        	maxn = p
        	minn = q
        else :
        	maxn = q
        	minn = p
        if maxn.val >root.val:
        	if minn.val <=root.val:
        		return root
        	else :
        		return self.lowestCommonAncestor(root.right, p, q)
        elif maxn.val == root.val:
        	return root
        else :
        	return self.lowestCommonAncestor(root.left, p, q)