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

22:T7e0,# Intuition\n\n\n# Approach\n\n\n# Complexity\n- Time complexity:\n\n\n- Space complexity:\n\n\n# Code\n```\n/**\n * Definition for a binary tree node.\n * struct TreeNode {\n * int val;\n * TreeNode *left;\n * TreeNode *right;\n * TreeNode() : val(0), left(nullptr), right(nullptr) {}\n * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}\n * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}\n * };\n */\nclass Solution {\nprivate:\n void Traverse(TreeNode* &root1,TreeNode* &root2)\n {\n if(root1==NULL || root2==NULL)return;\n if(root1->left && root2->left)root1->left->val=root1->left->val+root2->left->val;\n if(root1->right && root2->right)root1->right->val=root1->right->val+root2->right->val;\n if(root1->left && root2->left==NULL)root1->left->val=root1->left->val+0;\n if(root1->left==NULL && root2->left)\n {\n TreeNode* lnode=new TreeNode(root2->left->val);\n root1->left=lnode;\n }\n if(root1->right && root2->right==NULL)root1->right->val=root1->right->val+0;\n if(root1->right==NULL && root2->right)\n {\n TreeNode* rnode=new TreeNode(root2->right->val);\n root1->right=rnode;\n }\n Travers

Merge Two Binary Trees

Problem Description

Problem Description

Answer Panel