题解 | #合并二叉树#

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param t1 TreeNode类 
# @param t2 TreeNode类 
# @return TreeNode类
#
# 每次把两棵树的左子树和右子树合并，并返回根节点
# 需要判断左子节点和右子节点是否存在
class Solution:
    def mergeTrees(self , t1: TreeNode, t2: TreeNode) -> TreeNode:
        def merge(r1, r2):
            if r1 and r2:
                r1.val += r2.val
            elif not r1 and r2:
                return r2
            elif not r1 and not r2:
                return None
            
            if not r1.right:
                if r2.right:
                    r1.right = r2.right
            else:
                if r2.right:
                    r1.right = merge(r1.right, r2.right)
            
            if not r1.left:
                if r2.left:
                    r1.left = r2.left
            else:
                if r2.left:
                    r1.left = merge(r1.left, r2.left)
            
            return r1
        
        return merge(t1, t2)