101. Symmetric Tree (Easy) Microsoft Bloomberg LinkedIn 补Facebook

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.

Solution 1: Recursive O(n); O(h)

very similar with 100.Same Tree.

    public boolean isSymmetric(TreeNode root) {
        if (root == null) { //necessary
            return true;
        }
        return helper(root.left, root.right);
    }

    private boolean helper(TreeNode left, TreeNode right) {
        if (left == null || right == null) {
            return left == right;
        }

        if (left.val == right.val) {
            return helper(left.left, right.right) && helper(left.right, right.left);
        }
        return false;
    }

Solution 2: Iterative O(n); O(h)

    public boolean isSymmetricB(TreeNode root) {
        if (root == null) {
            return true;
        }

        Stack<TreeNode> stack = new Stack<>();
        stack.push(root.left);
        stack.push(root.right);
        while (!stack.empty()) {
            TreeNode n1 = stack.pop();
            TreeNode n2 = stack.pop();
            if (n1 == null && n2 == null) {
                continue;
            }
            if (n1 == null || n2 == null || n1.val != n2.val) {
                return false;
            }

            stack.push(n1.left);
            stack.push(n2.right);
            stack.push(n1.right);
            stack.push(n2.left);
        }
        return true;
    }