weekly contest 36的第一题。经典的递归吧，AC率高达70%+，难道大家都很懂递归吗。。我花了大概一个小时才AC。不过做出来了还是蛮开心的，证明我的思路是对的，只有在t1 t2都不为null的时候才不建树，其余情况直接return。其实这个buildTree的过程我模仿的是Serialize and deserialize a binary tree里面的那种方法。

    int value;
    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        TreeNode res = dfs(t1, t2);
        return res;
    }
    private TreeNode dfs(TreeNode t1, TreeNode t2) {
        if (t1 == null && t2 == null) {
            return null;
        }
        if (t1 != null && t2 != null) {
            value = t1.val + t2.val;
        } else if (t2 != null) {
            return t2;
        } else {
            return t1;
        }
        TreeNode node = new TreeNode(value);
        node.left = mergeTrees(t1.left, t2.left);
        node.right = mergeTrees(t1.right, t2.right);
        return node;
    }

我的if其实有点多，其实可以在所有的return之后再构造node的value。
漂亮的写法：

  public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        if (t1 == null)
            return t2;
        if (t2 == null)
            return t1;
        t1.val += t2.val;
        t1.left = mergeTrees(t1.left, t2.left);
        t1.right = mergeTrees(t1.right, t2.right);
        return t1;
    }
}