LeetCode刷题实战235：二叉搜索树的最近公共祖先
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).”
示例
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。
示例 2:
输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
输出: 2
解释: 节点 2 和节点 4 的最近公共祖先是 2, 因为根据定义最近公共祖先节点可以为节点本身。
说明:
所有节点的值都是唯一的。
p、q 为不同节点且均存在于给定的二叉搜索树中。
解题
如果两个节点值都小于根节点，说明他们都在根节点的左子树上，我们往左子树上找
如果两个节点值都大于根节点，说明他们都在根节点的右子树上，我们往右子树上找
如果一个节点值大于根节点，一个节点值小于根节点，说明他们他们一个在根节点的左子树上一个在根节点的右子树上，那么根节点就是他们的最近公共祖先节点。
递归法
class Solution(object):
def lowestCommonAncestor(self, root, p, q):
"""
:type root: TreeNode
:type p: TreeNode
:type q: TreeNode
:rtype: TreeNode
"""
if root.val < p.val and root.val < q.val:
return self.lowestCommonAncestor(root.right, p, q)
elif root.val > p.val and root.val > q.val:
return self.lowestCommonAncestor(root.left, p, q)
else:
return root
def lowestCommonAncestor(self, root, p, q):
"""
:type root: TreeNode
:type p: TreeNode
:type q: TreeNode
:rtype: TreeNode
"""
while root:
if root.val > max(p.val, q.val):
root = root.left
elif root.val < min(p.val, q.val):
root = root.right
else:
return root
return root