Leetcode617. 合併二叉樹

• 題目要求：

• 思路：

  • 修改Tree1，把Tree1和Tree2節點的和直接賦給Tree1
  • 遞歸
  • 判斷，若是t1和t2都不爲空，那麼把t2與t1的和賦給t1，而後遞歸調用當前的函數，把t1和t2的左節點做爲函數的參數，再把遞歸調用的整個值賦給t1的左節點（見代碼），右節點也是同樣
  • 最後把t1或t2返回
• 核心代碼：

if 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 or t2

• 完整代碼：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def mergeTrees(self, t1: TreeNode, t2: TreeNode) -> TreeNode:
        if 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 or t2