题干：

代码（递归实现）：

TreeNode* mergeTrees(TreeNode* root1, TreeNode* root2) {//前序好理解，直接将树覆盖到另一个上面if(root1 == NULL)return root2;//当前遍历节点为空的话就让另一个的值覆盖过来if(root2 == NULL)return root1;root1 -> val += root2 -> val;root1 -> left = mergeTrees(root1 -> left, root2 -> left);root1 -> right = mergeTrees(root1 -> right, root2 -> right);return root1;}

TreeNode* mergeTrees(TreeNode* root1, TreeNode* root2){//重新定义一棵树if(root1 == NULL)return root2;if(root2 == NULL)reutrn root1;TreeNode* node = new TreeNode(0);node -> left = mergeTrees(root1-> left, root2->left);node -> right = mergeTrees(root1-> right, root2-> right);return node;

代码（迭代实现）：在判断二叉树是否对称时涉及到了判断两个子树，用到了队列，那么处理两个树就使用队列。

TreeNode* mergeTrees(TreeNode* root1, TreeNode* root2) {//这里也是将树2覆盖到树1上if(root1 == NULL)return root2;if(root2 == NULL)return root1;queue<TreeNode*>que;if(root1 != NULL)que.push(root1);if(root2 != NULL)que.push(root2);while(!que.empty()){TreeNode* node1 = que.front();que.pop();TreeNode* node2 = que.front();que.pop();node1-> val += node2->val;//单层处理逻辑if(node1->left && node2-> left){que.push(node1->left);que.push(node2->left);//如果树1的左右非空，则分别与数2左右一并进入迭代逻辑循环}if(node1-> right && node2->right){que.push(node1->right);que.push(node2->right);//同样}if(!node1->left && node2->left){//都是判断树1左右是否空，空就让树2的左右覆盖，不会考虑树2node1->left = node2->left;}if(!node1->right && node2->right){node1->right = node2->right;}}return root1;}