// 如果我们找到了p，直接进行返回，那如果下面就是q呢？ 其实这没有影响，但是还是要多考虑一下
if (!root || root === p || root === q) return root;
const left = lowestCommonAncestor(root.left, p, q); // 去左边找，我们期望返回找到的节点
const right = lowestCommonAncestor(root.right, p, q); // 去右边找，我们期望返回找到的节点
if (!left) return right; // 左子树找不到，返回右子树
if (!right) return left; // 右子树找不到，返回左子树
return root; // 左右子树分别有一个，则返回root
如果没有明白的话，请多花时间消化一下
关键点解析
用递归的思路去思考树
代码
代码支持： JavaScript， Python3
JavaScript Code:
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @param {TreeNode} p
* @param {TreeNode} q
* @return {TreeNode}
*/
var lowestCommonAncestor = function (root, p, q) {
if (!root || root === p || root === q) return root;
const left = lowestCommonAncestor(root.left, p, q);
const right = lowestCommonAncestor(root.right, p, q);
if (!left) return right; // 左子树找不到，返回右子树
if (!right) return left; // 右子树找不到，返回左子树
return root; // 左右子树分别有一个，则返回root
};
Python Code:
# 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':
if not root or root == p or root == q:
return root
left = self.lowestCommonAncestor(root.left, p, q)
right = self.lowestCommonAncestor(root.right, p, q)
if not left:
return right
if not right:
return left
else:
return root
