101. Symmetric对称 Tree

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1
   / 
  2   2
 /  / 
3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

    1
   / 
  2   2
      
   3    3

Note:
Bonus points if you could solve it both recursively and iteratively. 

 

需要左右对称，解题思路和Same Tree类似。

如果拿掉root结点的话，逻辑其实相当于是比较root.left和root.right是否same tree类似的逻辑。只是需要把之前的left.left和right.right比较，left.right和right.left进行比较。

 

 public bool IsSymmetric(TreeNode root)
        {
            return IsMirror(root?.left, root?.right);
        }

        public bool IsMirror(TreeNode left, TreeNode right)
        {
            bool flag;
            if (left == null && right == null)
            {
                flag = true;
            }
            else if (left == null || right == null)
            {
                flag = false;
            }
            else
            {
                if (left.val == right.val)
                {
                    flag = IsMirror(left.left, right.right) && IsMirror(left.right, right.left);
                }
                else
                {
                    flag = false;
                }
            }

            return flag;
        }