<p>You are given two binary trees <code>root1</code> and <code>root2</code>.</p>

<p>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.</p>

<p>Return <em>the merged tree</em>.</p>

<p><strong>Note:</strong> The merging process must start from the root nodes of both trees.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/05/merge.jpg" style="width: 600px; height: 163px;" />
<pre>
<strong>Input:</strong> root1 = [1,3,2,5], root2 = [2,1,3,null,4,null,7]
<strong>Output:</strong> [3,4,5,5,4,null,7]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root1 = [1], root2 = [1,2]
<strong>Output:</strong> [2,2]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in both trees is in the range <code>[0, 2000]</code>.</li>
	<li><code>-10<sup>4</sup> &lt;= Node.val &lt;= 10<sup>4</sup></code></li>
</ul>

/**
 * 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) {
        // Si ambos árboles son nulos, retorna nullptr
        if (!root1 && !root2) return nullptr;

        // Crea el nodo raíz para el árbol resultante
        TreeNode* ans = new TreeNode();
        
        // Llama al método merge para realizar la fusión de los árboles
        merge(root1, root2, ans);
        
        return ans;
    }

    void merge(TreeNode* n1, TreeNode* n2, TreeNode* ans) {
        // Si ambos nodos son nulos, no hay nada que fusionar
        if (!n1 && !n2) return;

        // Sumar los valores de n1 y n2 si existen, sino usar 0
        ans->val = (n1 ? n1->val : 0) + (n2 ? n2->val : 0);

        // Si existe al menos uno de los nodos hijos izquierdos, crear nuevo nodo en ans->left
        if (n1 || n2) {
            if (n1 && n1->left || n2 && n2->left) {
                ans->left = new TreeNode();
                merge(n1 ? n1->left : nullptr, n2 ? n2->left : nullptr, ans->left);
            }

            // Si existe al menos uno de los nodos hijos derechos, crear nuevo nodo en ans->right
            if (n1 && n1->right || n2 && n2->right) {
                ans->right = new TreeNode();
                merge(n1 ? n1->right : nullptr, n2 ? n2->right : nullptr, ans->right);
            }
        }
    }
};

34600