LeetCode 617. Merge Two Binary Trees

题目链接：https://leetcode.com/problems/merge-two-binary-trees/description/

题目解析：将T1作为模板，把T2合进来。容易想到的是用递归的方法遍历子树，主要分成三种情况——

• T1和T2当前节点都存在，那么两者值相加即可，继续分别递归左右子树；
• T1不存在，T2存在，那么把T1当前空节点赋值为T2当前节点即可，以T2当前节点为根结点的子树自然会合并到T1上；
• T1存在，T2不存在，那么只要返回即可，不用进行任何操作。

代码如下：20ms Accepted beating 100%

/**
 * 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 && t2 != NULL)
        {
            t1->val = t1->val + t2->val;
            t1->right = mergeTrees(t1->right, t2->right);
            t1->left = mergeTrees(t1->left, t2->left);
            return t1;
        }
        else if (t2 != NULL)
        {
            t1 = t2;
            return t1;
        }
        else
            return t1;
    }
};

static const auto ____ = [] () {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    return nullptr;
}();