Problem

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

思路

• 看当前这颗树是否对称，需要看，他的左子树和右子树是否对称
• 所以创建一个函数检测给定的两个树是否对称
  • 根节点的值必须相同
  • A左和B右相同， A右和B左相同

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

    private boolean isSymmetric(TreeNode left, TreeNode right) {
        if (left == null || right == null) {
            return left == right;
        }
        if (left.val != right.val) return false;
        return isSymmetric(left.left, right.right) && isSymmtric(left.right, right.left);
    }

stack 版本

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

        TreeNode l = root.left;
        TreeNode r = root.right;
        if (l == null && r == null) return true;

        Stack<TreeNode> stack = new Stack<>();
        if (l != null && r != null && l.val == r.val) {
            stack.push(l);
            stack.push(r);
        } else return false;

        while (!stack.isEmpty()) {
            if (stack.size() % 2 != 0) return false;
            TreeNode right = stack.pop();
            TreeNode left = stack.pop();

            if (left.val != right.val) return false;
            if (left.left != null) {
                if (right.right == null) return false;
                stack.push(left.left);
                stack.push(right.right);
            } else if (right.right != null) {
                return false;
            }

            if (left.right != null) {
                if (right.left == null) return false;
                stack.push(left.right);
                stack.push(right.left);
            } else if (right.left != null) {
                return false;
            }

        }
        return true;
    }