C++ 解法, 执行用时: 2ms, 内存消耗: 356KB, 提交时间: 2020-12-20

class Solution {
  public:
    /**
     *
     * @param t1 TreeNode类
     * @param t2 TreeNode类
     * @return TreeNode类
     */
    TreeNode *mergeTrees(TreeNode *t1, TreeNode *t2) {
        // write code here
        if (t1 == nullptr && t2 == nullptr)
            return nullptr;
        else if (t1 == nullptr)
            return t2;
        else if (t2 == nullptr)
            return t1;
        else {
            TreeNode *t = new TreeNode(t1->val + t2->val);
            t->left = mergeTrees(t1->left, t2->left);
            t->right = mergeTrees(t1->right, t2->right);
            return t;
        }
    }
};

C++ 解法, 执行用时: 2ms, 内存消耗: 380KB, 提交时间: 2021-07-18

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param t1 TreeNode类 
     * @param t2 TreeNode类 
     * @return TreeNode类
     */

    
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        // write code here
        if(!t1) return t2;
        if(!t2) return t1;
        
        t1->val += t2->val;
        t1->left = mergeTrees(t1->left, t2->left);       
        t1->right = mergeTrees(t1->right, t2->right);
        return t1;
    }
};

C++ 解法, 执行用时: 2ms, 内存消耗: 380KB, 提交时间: 2021-07-13

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param t1 TreeNode类 
     * @param t2 TreeNode类 
     * @return TreeNode类
     */
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        if(!t1 && !t2) return nullptr;
        TreeNode *tmp = new TreeNode(-1);
        tmp->val = (t1?t1->val:0) + (t2?t2->val:0);
        tmp->left = mergeTrees(t1?t1->left:nullptr, t2?t2->left:nullptr);
        tmp->right = mergeTrees(t1?t1->right:nullptr, t2?t2->right:nullptr);
        return tmp;
    }
};

C++ 解法, 执行用时: 2ms, 内存消耗: 380KB, 提交时间: 2021-07-13

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */

class Solution {
public:
  
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        if(!t1)return t2;
        if(!t2)return t1;
        t1->val+=t2->val;
        t1->left=mergeTrees(t1->left, t2->left);
        t1->right=mergeTrees(t1->right, t2->right);
        return t1;
    }
};

C++ 解法, 执行用时: 2ms, 内存消耗: 384KB, 提交时间: 2021-07-10

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param t1 TreeNode类 
     * @param t2 TreeNode类 
     * @return TreeNode类
     */
    TreeNode* dfs(TreeNode* t1, TreeNode* t2)
    {
        TreeNode* ans;
        if(t1==nullptr&&t2==nullptr)
            return nullptr;
        if(t1==nullptr)
            ans=new TreeNode(t2->val);
        else if(t2==nullptr)
            ans=new TreeNode(t1->val);
        else
            ans=new TreeNode(t2->val+t1->val);
        ans->left=dfs((t1==nullptr)?nullptr:t1->left,(t2==nullptr)?nullptr:t2->left);
        ans->right=dfs((t1==nullptr)?nullptr:t1->right,(t2==nullptr)?nullptr:t2->right);
        return ans;
    }
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        // write code here
        return dfs(t1,t2);
    }
};