動機
善用binary search tree,所以才是easy
Problem
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).”
Example 1:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8Output: 6Explanation: The LCA of nodes 2 and 8 is 6.
Example 2:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4Output: 2Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.
Example 3:
Input: root = [2,1], p = 2, q = 1Output: 2
Constraints:
- The number of nodes in the tree is in the range
[2, 105]
. -109 <= Node.val <= 109
- All
Node.val
are unique. p != q
p
andq
will exist in the BST.
Sol
p,q其實就是範圍,所以就照BTS的方式去找就好
class Solution:
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
if p.val > q.val:
return self.lowestCommonAncestor(root,q,p)
if not root:
return None
elif p.val <= root.val <= q.val:
return root
elif q.val < root.val:
return self.lowestCommonAncestor(root.left,p,q)
else:
return self.lowestCommonAncestor(root.right,p,q)