题解 | #合并二叉树#

- 递归，直接在t1上修改
 
/**
 * 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;   //如果t2节点不存在，则用t1
        t1->val = t1->val + t2->val;  //t1 t2都存在，t1的值更新为两节点值之和
        t1->left = mergeTrees(t1->left, t2->left);
        t1->right = mergeTrees(t1->right, t2->right);
        return t1;
    }
};