[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



思路：我把t1作为最后的结果返回。同时递归，t1和t2 把t2的值加到t1上去，如果t1为空了而t2不为空
，那么我们把t2接到t1上去，同理,如果t2为空了,我们把t1接到t1上去。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        if (t1==null&&t2==null)
            return null;
        if (t1==null&&t2!=null)
            return t2;
        if (t1!=null&&t2==null)
            return t1;
        t1.val+=t2.val;
        t1.left = mergeTrees(t1.left,t2.left);
        t1.right = mergeTrees(t1.right,t2.right);
        return t1;
    }
}