题解 | #合并二叉树#

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

class Solution {
  public:
    /**
     *
     * @param t1 TreeNode类
     * @param t2 TreeNode类
     * @return TreeNode类
     */
  	//采用中序遍历的思路两个树进行遍历
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        // write code here
        if (t1 == nullptr) { //当t1存在空节点时，将t2的节点赋给t1
            t1 = t2;
            return t1;
        }else if(t2 == nullptr){//当t2存在空节点时，将t1的节点直接返回
            return t1;
        }
	  	//中序操作
        t1->val = t1->val+t2->val;
	  	//左子树操作
        t1->left = mergeTrees(t1->left, t2->left);
	  	//右子树操作
        t1->right = mergeTrees(t1->right, t2->right);
        return t1;
    }
};