## 题目描述

Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not.

You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of new tree.

Example 1:

``````Input:
Tree 1                     Tree 2
1                         2
/ \                       / \
3   2                     1   3
/                           \   \
5                             4   7
Output:
Merged tree:
3
/ \
4   5
/ \   \
5   4   7``````

Note: The merging process must start from the root nodes of both trees.

## 解法一

### 思路

• 若n1和n2都存在，则只需要保留其中一个结点（如n1），将另一结点的值加到此结点上即可（如`n1.val += n2.val`）。
• 若n1或n2任一不存在，则合并后的二叉树对应位置上的结点就是存在的那个了。
• 若n1和n2都不存在，则合并后仍不存在。

### Python

``````class Solution:
def mergeTrees(self, t1, t2):
"""
:type t1: TreeNode
:type t2: TreeNode
:rtype: TreeNode
"""
if t1 is not None and t2 is not None:
t1.left = self.mergeTrees(t1.left, t2.left)
t1.right = self.mergeTrees(t1.right, t2.right)
t1.val += t2.val
return t1
return t1 if t2 is None else t2``````

### Java

``````/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
if (t1 != null && t2 != null) {
t1.left = mergeTrees(t1.left, t2.left);
t1.right = mergeTrees(t1.right, t2.right);
t1.val += t2.val;
return t1;
}
return t1 == null ? t2 : t1;
}
}``````

### C++

``````/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode *mergeTrees(TreeNode *t1, TreeNode *t2) {
if (t1 != nullptr && t2 != nullptr) {
t1->left = mergeTrees(t1->left, t2->left);
t1->right = mergeTrees(t1->right, t2->right);
t1->val += t2->val;
return t1;
}
return t1 == nullptr ? t2 : t1;
}
};``````