617. 合并二叉树

递归合并

/**
 * 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) {}
 * };
 */
#define solve(a,b) ((b)+=((a)?(a->val):0))
#define judge_left(a) ((a)&&(a->left))
#define judge_right(a) ((a)&&(a->right))
#define get_left(a) (judge_left(a)?(a)->left : NULL)
#define get_right(a) (judge_right(a)?(a)->right : NULL)

class Solution {
public:
    TreeNode* mergeTrees(TreeNode* root1, TreeNode* root2) {
        TreeNode *root = NULL;
        if (root1 || root2)
        {
            int val = 0;
            solve(root1, val);
            solve(root2, val);
            root = new TreeNode(val, NULL, NULL);
            if (judge_left(root1) || judge_left(root2))
                root->left = mergeTrees(get_left(root1), get_left(root2));
            if (judge_right(root1) || judge_right(root2))
                root->right = mergeTrees(get_right(root1), get_right(root2));
        }
        return root;
    }
};