Q;
中最近公共祖先的定义为:“对于有根树 T 的两个结点 p、q,最近公共祖先表示为一个结点 x,满足 x 是 p、q 的祖先且 x 的深度尽可能大(一个节点也可以是它自己的祖先)。”
例如,给定如下二叉搜索树: root = [6,2,8,0,4,7,9,null,null,3,5]
示例 1:
输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8 输出: 6 解释: 节点2
和节点8
的最近公共祖先是6。
链接:https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-search-tree/description/
思路:首先二叉搜索树是二叉树的子集,因此这道题完全可以按照 二叉树的最近公共祖先 来做,但是问题就是这样没有用到二叉搜索树的条件,也就是左叶子的值小于根节点,右叶子的值大于根节点,因此这个题两种解决方法
代码:
完全按照二叉树的思想做:
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def lowestCommonAncestor(self, root, p, q):
"""
:type root: TreeNode
:type p: TreeNode
:type q: TreeNode
:rtype: TreeNode
"""
if(root is None or root==p or root==q):
return root
left=self.lowestCommonAncestor(root.left,p,q)
right=self.lowestCommonAncestor(root.right,p,q)
if left and right :
return root
if not left :
return right
if not right :
return left
return None
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def lowestCommonAncestor(self, root, p, q):
"""
:type root: TreeNode
:type p: TreeNode
:type q: TreeNode
:rtype: TreeNode
"""
if not root:
return(root)
if p.val < root.val and q.val < root.val:
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)