難度：簡單
題目描述：
思路總結：
難者不會，做過就不難。利用二叉搜尋樹的性質，左子樹《根節點《右子樹。根據經驗可以得到，如果p和q不再當前結點的某一子樹里，（二叉搜素樹性質，同時大於，或者同時小於。）就是最近公共祖先。
題解一：（遞迴）
# 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'
:#思路：
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
題解二：（迭代）
# 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'
:#思路：
stack =
[root]
while stack:
cur = stack.pop(
)if p.val < cur.val and q.val < cur.val:
elif p.val > cur.val and q.val > cur.val:
else
:return cur
#改進後
# 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'
:#思路：
node = root
while node:
if p.val < node.val and q.val < node.val:
node = node.left
elif p.val > node.val and q.val > node.val:
node = node.right
else
:return node
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