給定兩個二叉樹，想象當你將它們中的一個覆蓋到另一個上時，兩個二叉樹的一些節點便會重疊。

你需要將他們合併為一個新的二叉樹。合併的規則是如果兩個節點重疊，那麼將他們的值相加作為節點合併後的新值，否則不為 NULL 的節點將直接作為新二叉樹的節點。

示例 1:

輸入: 
    Tree 1                     Tree 2                  
          1                         2                             
         / \                       / \                            
        3   2                     1   3                        
       /                           \ 
 
  \                      
      5                             4   7                  
輸出: 
合併後的樹:
         3
        / \
       4   5
      / \   \ 
     5   4   7

注意: 合併必須從兩個樹的根節點開始。

我沒繞過來的點：當兩個數節點都有值的時候，需要返回兩個節點值的和；遞迴只有當兩者都有值的時候才能繼續往下進行。如果一旦某個地方只有一個值，那麼返回這個值（注意，此時這個值是樹的節點，和連結串列一樣，是包含它的兩個子節點的，比如兩個左下連結串列的樹和一個右下連結串列的樹）

class Solution {
    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        TreeNode root = null;
        if(t1 == null && t2 == null)
            return root;
        if(t1 == null && t2 != null)
           return t2;                   
        if(t1 != null && t2 == null
 
)
            return t1;            

        root = new TreeNode(t1.val + t2.val);
        root.left = mergeTrees(t1.left,t2.left);
        root.right = mergeTrees(t1.right,t2.right);       

        return root;
    }
}

class Solution {
    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        if(t1 == null && t2 == null)
            return null;

        if(t1 != null && t2 != null){
            t1.val += t2.val; 
            t1.left = mergeTrees(t1.left,t2.left);
            t1.right = mergeTrees(t1.right,t2.right);
        }
        if(t1 == null && t2 != null){
            t1 = new TreeNode(t2.val);
            t1.left = mergeTrees(null,t2.left);
            t1.right = mergeTrees(null,t2.right);
        }
        if(t1 != null && t2 == null){
            t1.left = mergeTrees(t1.left,null);
            t1.right = mergeTrees(t1.right,null);
        }

        return t1;
    }
}

犯的錯誤：沒有新增t1.left = …和t1.right=…

class Solution {
    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        TreeNode root = null;
        if(t1 == null && t2 == null)
            return root;
        if(t1 == null && t2 != null){
            root = new TreeNode(t2.val);
            root.left = mergeTrees(null,t2.left);
            root.right = mergeTrees(null,t2.right);
        }
        if(t1 != null && t2 == null){
             root = new TreeNode(t1.val);
             root.left = mergeTrees(t1.left,null);
             root.right = mergeTrees(t1.right,null);
        }
        if(t1 != null && t2 != null){
            root = new TreeNode(t1.val + t2.val);
            root.left = mergeTrees(t1.left,t2.left);
            root.right = mergeTrees(t1.right,t2.right);
        }

        return root;
    }
}

class Solution {
    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        TreeNode root = null;
        if(t1 == null && t2 == null)
            return root;
        if(t1 == null && t2 != null){
            root = t2;           
        }
        if(t1 != null && t2 == null){
             root = t1;            
        }
        if(t1 != null && t2 != null){
            root = new TreeNode(t1.val + t2.val);
            root.left = mergeTrees(t1.left,t2.left);
            root.right = mergeTrees(t1.right,t2.right);
        }

        return root;
    }
}