题目：
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______
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___2__ ___8__
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0 _4 7 9
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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.
链接： http://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/
一刷，思想是判断当前节点的值
1.与输入的两值之一相等，查找另一值是否在树里，是则当前值是LCA，否则返回None
2.当前值在两值中间，查找输入两值是否在树里，都在则返回当前值，否则返回None
3.当前值比最小值小或比最大值大，更新当前值
问题，若输入查找的值是None，输出None还是另外一值呢？
1 class Solution(object): 2 @staticmethod 3 def find(root, p): 4 if not p: 5 return False 6 cur = root 7 while root: 8 if root.val == p.val: 9 return True 10 if root.val < p.val: 11 root = root.right 12 else: 13 root = root.left 14 return False 15 16 def lowestCommonAncestor(self, root, p, q): 17 """ 18 :type root: TreeNode 19 :type p: TreeNode 20 :type q: TreeNode 21 :rtype: TreeNode 22 """ 23 if not root: 24 return root 25 if not p or not q: 26 return None 27 small = p if p.val < q.val else q 28 large = p if p.val >= q.val else q 29 30 cur = root 31 while cur: 32 if cur.val == small.val: 33 return cur if Solution.find(cur, large) else None 34 if cur.val == large.val: 35 return cur if Solution.find(cur, small) else None 36 if small.val < cur.val < large.val: 37 if Solution.find(cur, small) and Solution.find(cur, large): 38 return cur 39 else: 40 return None 41 cur = cur.right if cur.val < small.val else cur.left 42 return None
如果没有找到node可以返回另一个的话，二刷可以试一下recursive解法。