题目描述

Given two nodes of a binary tree p and q, return their lowest common ancestor (LCA).

Each node will have a reference to its parent node. The definition for Node is below:

class Node {
public int val;
public Node left;
public Node right;
public Node parent;
}
According to the definition of LCA on Wikipedia: “The lowest common ancestor of two nodes p and q in a tree T is the lowest node that has both p and q as descendants (where we allow a node to be a descendant of itself).”

样例

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.

Input: root = [1,2], p = 1, q = 2
Output: 1

3.
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 and q exist in the tree.

"""
# Definition for a Node.
class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None
        self.parent = None
"""

class Solution:
    def lowestCommonAncestor(self, p: 'Node', q: 'Node') -> 'Node':
        x, y = p, q
        while x != y:
            x = x.parent if x.parent else q
            y = y.parent if y.parent else p
        return x

算法2

"""
# Definition for a Node.
class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None
        self.parent = None
"""

class Solution(object):
    def lowestCommonAncestor(self, p, q):
        """
        :type node: Node
        :rtype: Node
        """
        result = set()
        while p:
            result.add(p.val)
            p = p.parent
        while q:
            if q.val in result:
                return q
            q = q.parent
        return None