-
Notifications
You must be signed in to change notification settings - Fork 25
Expand file tree
/
Copy path236_LowestCommonAncestorOfABinaryTree.py
More file actions
50 lines (38 loc) · 1.47 KB
/
Copy path236_LowestCommonAncestorOfABinaryTree.py
File metadata and controls
50 lines (38 loc) · 1.47 KB
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
# coding: utf8
"""
题目链接: https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/description.
题目描述:
Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
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).”
_______3______
/ \
___5__ ___1__
/ \ / \
6 _2 0 8
/ \
7 4
For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5,
since a node can be a descendant of itself according to the LCA definition.
"""
# 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 in (None, p, q):
return root
left = self.lowestCommonAncestor(root.left, p, q)
right = self.lowestCommonAncestor(root.right, p, q)
if left and right:
return root
return left if left else right