二叉树的最近公共祖先

还可以再优化，使用递归两步解决，第一步找出p的所有父节点，第二步从p的最后一个父节点开始依次找q

# 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: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':

        def commonAncestor(root, q, res):
            if root is None:
                return False
            if root.val == q.val:
                res.append(root)
                return True
            if root.left is not None:
                if commonAncestor(root.left, p, res):
                    res.append(root)
                    return True
            if root.right is not None:
                if commonAncestor(root.right, p, res):
                    res.append(root)
                    return True

        def commonAncestorSplit(root, q):
            if root is None:
                return False
            if root.val == q.val:
                return True
            if root.left is not None:
                if commonAncestorSplit(root.left, q):
                    return True
            if root.right is not None:
                if commonAncestorSplit(root.right, q):
                    return True
            return False

        res = []
        commonAncestor(root, p, res)
        for r in res:
            if commonAncestorSplit(r, q):
                return r