思路:递归求解某一棵树的左右节点,返回
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) { if(t1 ==NULL) { return t2; } if (t2 == NULL) { return t1; } t1->val+= t2->val; t1->left = mergeTrees(t1->left,t2->left); t1->right = mergeTrees(t1->right,t2->right); return t1; } };