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
#
class Solution:
    def mergeTrees(self , t1: TreeNode, t2: TreeNode) -> TreeNode:
        # write code here
        if not t1:
            return t2
        if not t2:
            return t1

        t = TreeNode(t1.val+t2.val)
        t.left = self.mergeTrees(t1.left, t2.left)
        t.right = self.mergeTrees(t1.right, t2.right)

        return t

这道题目考察同时遍历两个二叉树的情况，由于都是相同位置合并，所以是同步进行遍历的，采用递归很容易得到合并的代码.