LeetCode 617. Merge Two Binary Tree （合并两个二叉树）

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.

题目标题：Tree

　　这道题目给了我们两个二叉树，要我们合并两个二叉树，合并的树的每一个点的值等于两个二叉树相对位置的点的值的合。利用recursively call来实现，我们来分析一下。对于每一个新的点的值，我们需要做的就是把两个树中的同样位置的点的值相加。然后recursively来继续代入mergeTrees，左边的点，就代入同样位置两个点的左边。右边的点就代入同样位置的两个点的右边，直到代入得两个点都是null，就停止代入，return回去。那么对于每一个新的点，有三种情况：1- 两个点都是null，就直接return； 2- 两个点都不是null，直接相加；3- 两个点其中有一个点是null，那么就取另外一个点的值。需要注意的是，对于每一个新的点，如果代入的两个点其中一个是null的话，那么这个null的点的 .left 和.right 是error。所以要先initial 一下。

Java Solution:

Runtime beats 60.24%

完成日期：06/29/2017

关键词：Tree

关键点：利用recursively来求每一个新的点，以及这个点的左右child

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution 
11 {
12     public TreeNode mergeTrees(TreeNode t1, TreeNode t2) 
13     {
14         TreeNode root;
15         TreeNode left_1 = null, left_2 = null;
16         TreeNode right_1 = null, right_2 = null;
17         
18         if(t1 == null && t2 == null)
19             return null;
20         else if(t1 != null && t2 != null)
21         {
22             root = new TreeNode(t1.val + t2.val);
23             left_1 = t1.left;
24             left_2 = t2.left;
25             right_1 = t1.right;
26             right_2 = t2.right;
27         }
28         else if(t1 != null && t2 == null)
29         {
30             root = new TreeNode(t1.val);
31             left_1 = t1.left;
32             right_1 = t1.right;
33         }
34         else
35         {
36             root = new TreeNode(t2.val);
37             left_2 = t2.left;
38             right_2 = t2.right;
39         }
40         
41         
42         root.left = mergeTrees(left_1, left_2);
43         root.right = mergeTrees(right_1, right_2);
44         
45             
46         return root;
47     }
48 }