1126. 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
Input:
Tree 1 Tree 2
1 2
/ \ / \
3 2 1 3
/ \ \
5 4 7
Output:
Merged tree:
3
/
4 5
/ \ \
5 4 7
Notice
The merging process must start from the root nodes of both trees.

思路：
二叉樹遞迴。
注意：

1. Tree * node = NULL; //記得這裡要初始化為NULL, 否則當t1, t2都是NULL時，返回unknown。
   程式碼如下：

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param t1: the root of the first tree
     * @param t2: the root of the second tree
     * @return: the new binary tree after merge
     */
    TreeNode * mergeTrees(TreeNode * t1, TreeNode * t2) {
        TreeNode * node = NULL;
        if (t1 && t2) {
            node = new TreeNode(t1->val + t2->val);
            node->left = mergeTrees(t1->left, t2->left);
            node->right = mergeTrees(t1->right, t2->right);
        } else if (t1 && !t2) {
            node = new TreeNode(t1->val);
            node->left = mergeTrees(t1->left, NULL);
            node->right = mergeTrees(t1->right, NULL);
        } else if (!t1 && t2) {
            node = new TreeNode(t2->val);
            node->left = mergeTrees(NULL, t2->left);
            node->right = mergeTrees(NULL, t2->right);
        }     
        return node;
    }
};