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

合并两棵二叉树，对应位置都存在结点则将它们的值加起来，若只有一棵树有节点则用该结点替代。

Answer：

对于二叉树最常使用的是递归方法。如果t1结点为空，则返回t2结点；若t2结点为空就返回t1结点；否则，建立新节点，值为t1与t2之和，然后递归的调用函数，构造该节点的左右孩子结点。

class Solution {
public:
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        if(!t1)
            return t2;
        if(!t2)
            return t1;
        TreeNode *n=new TreeNode(t1->val+t2->val);
        n->left=mergeTrees(t1->left,t2->left);
        n->right=mergeTrees(t1->right,t2->right);
        return n;
    }
};

Merge Two Binary Trees[LeetCode]