Leetcode 617. Merge Two Binary Trees

Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not.

You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of new tree.

Example 1:
Input:
Tree 1 Tree 2
1 2
/ \ / \
3 2 1 3
/ \ \
5 4 7
Output:
Merged tree:
3
/ \
4 5
/ \ \
5 4 7
Note: The merging process must start from the root nodes of both trees.

一道很简单的有关树的遍历的题。题意就是要合并两棵树，相同位置的节点的值相加。同样是用递归来做，自己的想法还是显得复杂许多。最后面给出了大神写的代码，只有几行代码简洁又优雅。
也很好理解，若有一个树为空，则直接接上另外一个树的余下部分，若两个树都不为空，则把值相加。

class Solution {
public:
    TreeNode* result;
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        TreeNode * res;
        travel(result,t1,t2);

        return result;
    }
    void travel(TreeNode* &res,TreeNode *x,TreeNode *y) {
        int val1=0, val2=0;

        if (x == NULL&&y==NULL) {
            return;
        } 
        if (x == NULL) {


            res = new TreeNode(y->val);
            travel(res->left, NULL, y->left);
            travel(res->right, NULL, y->right);
        }else if (y == NULL) {

            res = new TreeNode(x->val);
            travel(res->left, x->left, NULL);
            travel(res->right, x->right, NULL);
        }
        else {
            res = new TreeNode(x->val+y->val);
            travel(res->left, x->left, y->left);
            travel(res->right, x->right, y->right);
        }


    }
};

class Solution {
public:
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        if (!t1) return t2;
        if (!t2) return t1;

        TreeNode* node = new TreeNode(t1->val + t2->val);
        node->left = mergeTrees(t1->left, t2->left);
        node->right = mergeTrees(t1->right, t2->right);
        return node;
    }
};