# Leetcode Merge-Two-Binary-Trees

#### 2022-07-16 09:11

> ##### Algorithms:
>
> #algorithm #DFS #DFS_inorder
>
> ##### Data structures:
>
> #DS #binary_tree
>
> ##### Difficulty:
>
> #coding_problem #difficulty-easy
>
> ##### Additional tags:
>
> #leetcode
>
> ##### Revisions:
>
> N/A

##### Related topics:

##### Links:

- [Link to problem](https://leetcode.com/problems/merge-two-binary-trees/)

---

### Problem

You are given two binary trees `root1` and `root2`.

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 the two trees 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 the new tree.

Return _the merged tree_.

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

#### Examples

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/02/05/merge.jpg)

**Input:** root1 = [1,3,2,5], root2 = [2,1,3,null,4,null,7]
**Output:** [3,4,5,5,4,null,7]

**Example 2:**

**Input:** root1 = [1], root2 = [1,2]
**Output:** [2,2]

#### Constraints

- The number of nodes in both trees is in the range `[0, 2000]`.
- `-104 <= Node.val <= 104`

### Thoughts

> [!summary]
> This is a #DFS problem, can be solved with recursion

DFS preorder recursion:

- Base case: one leaf is empty, return another
- add value from root2 to root1

- root1->left = recurse
- root1->right = recurse

- return root1

### Solution

```cpp
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
  TreeNode *mergeTrees(TreeNode *root1, TreeNode *root2) {
    // DFS preorder

    // if both are nullptr, return root2, which is a nullptr
    if (!root1) {
      return root2;
    } else if (!root2) {
      return root1;
    }

    root1->val += root2->val;

    root1->left = mergeTrees(root1->left, root2->left);
    root1->right = mergeTrees(root1->right, root2->right);

    return root1;
  }
};
```
