Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
According to thedefinition 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 allowa node to be a descendant of itself).”
Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]
Example 1:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of of nodes 5 and 1 is 3.
Example 2:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself
according to the LCA definition.
Note:
All of the nodes' values will be unique.
p and q are different and both values will exist in the binary tree.
分析：
不是搜索树。用的分治
注意tuple 可以这样 a,b = (a,b)
# 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, p, q):
"""
:type root: TreeNode
:type p: TreeNode
:type q: TreeNode
:rtype: TreeNode
"""
if not root:
return root
if root.val == p.val:
return p
if root.val == q.val:
return q
left = self.lowestCommonAncestor(root.right,p,q)
right = self.lowestCommonAncestor(root.left,p,q)
if not left:
return right
if not right:
return left
if left.val == p.val and right.val == q.val or left.val == q.val and right.val == p.val:
return root
# 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, p, q):
if root in (None, p, q): return root
left,right = (self.lowestCommonAncestor(kid, p, q)
for kid in (root.left, root.right))
return root if left and right else left or right