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.

Input: 
	Tree 1                     Tree 2                  
          1                         2                             
         / \                       / \                            
        3   2                     1   3                        
       /                           \   \                      
      5                             4   7                  
Output: 
Merged tree:
	     3
	    / \
	   4   5
	  / \   \ 
	 5   4   7

 

解法一：
以t1為合併的新樹，同時遍歷兩棵樹，遍歷節點都存在則新節點為兩節點值之和，若一方不存在返回另一個，都不存在返回Null。

/**
 * 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 = t1.val + t2.val;
        t1.left = mergeTrees(t1.left, t2.left);
        t1.right = mergeTrees(t1.right, t2.right);
        return t1;
    }
}

解法二：DFS遍歷合併

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    
    public void DFS(TreeNode t1, TreeNode t2, TreeNode ans) {
        if (t1 != null && t2 != null) {
            ans.val = t1.val + t2.val;
            if (t1.left != null || t2.left != null) {
                ans.left = new TreeNode(0);
                DFS(t1.left, t2.left, ans.left);
            }
            if (t1.right != null || t2.right != null) {
                ans.right = new TreeNode(0);
                DFS(t1.right, t2.right, ans.right);
            }
            return;
        } else if (t1 != null) {
            ans.val = t1.val;
            if (t1.left != null) {
                ans.left = new TreeNode(0);
                DFS(t1.left, null, ans.left);
            }
            if (t1.right != null) {
                ans.right = new TreeNode(0);
                DFS(t1.right, null, ans.right);
            }
        } else if (t2 != null) {
            ans.val = t2.val;
            if (t2.left != null) {
                ans.left = new TreeNode(0);
                DFS(null, t2.left, ans.left);
            }
            if (t2.right != null) {
                ans.right = new TreeNode(0);
                DFS(null, t2.right, ans.right);
            }
        }
    }
    
    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        TreeNode ans = null;
        if (t1 == null && t2 == null) {
            return ans;
        }
        ans = new TreeNode(0);
        DFS(t1, t2, ans);
        return ans;
    }