1. 题目

给定两个二叉树，想象当你将它们中的一个覆盖到另一个上时，两个二叉树的一些节点便会重叠。

你需要将他们合并为一个新的二叉树。合并的规则是如果两个节点重叠，那么将他们的值相加作为节点合并后的新值，否则不为 NULL 的节点将直接作为新二叉树的节点。

示例 1:输入: Tree 1                     Tree 2                  1                         2                             / \                       / \                            3   2                     1   3                        /                           \   \                      5                             4   7                  
输出: 
合并后的树:3/ \4   5/ \   \ 5   4   7
注意: 合并必须从两个树的根节点开始。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/merge-two-binary-trees
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

2. 递归解题

class Solution {
public:TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {if(!t1 && !t2)return NULL;TreeNode *node = new TreeNode(0);if(t1 && !t2){node->val = t1->val;node->left = t1->left;node->right = t1->right;}else if(!t1 && t2){node->val = t2->val;node->left = t2->left;node->right = t2->right;}else//两个树种节点都存在{node->val = t1->val + t2->val;node->left = mergeTrees(t1->left, t2->left);node->right = mergeTrees(t1->right, t2->right);}return node;}
};

class Solution {// c++ 2020.9.23
public:TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {if(!t1) return t2;if(!t2) return t1;auto l = mergeTrees(t1->left, t2->left);auto r = mergeTrees(t1->right, t2->right);t1->val += t2->val;t1->left = l;t1->right = r;return t1;}
};

class Solution: # py3def mergeTrees(self, t1: TreeNode, t2: TreeNode) -> TreeNode:if not t1:return t2elif not t2:return t1l = self.mergeTrees(t1.left, t2.left)r = self.mergeTrees(t1.right, t2.right)t1.val += t2.valt1.left = lt1.right = rreturn t1

100 ms 14.5 MB