class Solution:
def mergeTrees(self , t1: TreeNode, t2: TreeNode) -> TreeNode:
if not t1 and not t2:
return None
elif t1 and t2:
t1.val += t2.val
t1.left = self.mergeTrees(t1.left, t2.left)
t1.right = self.mergeTrees(t1.right, t2.right)
return t1
else:
if not t1:
return t2
else:
return t1
选择 t1 作为最终返回的二叉树
若 t1, t2 均不存在,则返回 None
若 t1, t2 均存在,则令 t1.val 与 t2.val 相加并存于 t1.val中,同时获取左右子树根节点,设为 t1 的左右子树后返回 t1
若 t1, t2 只存在其中一个,则返回存在的节点(成为上一递归的左/右子树)