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 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).”

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 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.

这个题目还是用Divide and Conquer的思想，然后分别在左右两边tree里面找是否有p，或者g，一旦有了，就分别往上面传，只有在left tree， right tree 各有一个点时，就是lowest common ancestor。

Time：O(n),   S: O(n)

Code：

class Solution:

    def LCA(self, root, p, g):

        if not root or root == p or root == g:

            return root

        left = self.LCA(root.left, p, g)

        right = self.LCA(root.right, p, g)

        if left and right:

            return root

        elif left:

            return left

        elif right:

            return right

        return 

=> 更简洁

class Solution:

    def LCA(self, root, p, g):

        if root in [None, p, g]: return root

        left, right = self.LCA(root.left, p, g), self.LCA(root.right, p, g)

        return root if left and right else left or right