题目要求：思路：修改Tree1，将Tree1和Tree2节点之和直接赋值给Tree1进行递归判断，如果t1和t2都不为空，则将t2和t1之和赋值给t1，而然后递归调用当前函数，将t1和t2的左节点作为函数的参数，然后将递归调用的整个值赋值给t1的左节点（见代码），右节点同理，最后将t1或t2返回到核心代码：ift1andt2:t1.val+=t2.valt1.left=self.mergeTrees(t1.left,t2.left)t1.right=self.mergeTrees(t1.right,t2.right)returnt1ort2completecode:#定义二叉树节点。#classTreeNode:#def__init__(self,x):#self.val=x#self.left=None#self.right=None类解决方案：defmergeTrees(self,t1:TreeNode,t2:TreeNode)->TreeNode:如果t1和t2:t1.val+=t2.valt1.left=self.mergeTrees(t1.left,t2.left)t1.right=self.mergeTrees(t1.right,t2.right)返回t1或t2
