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:

Example：

[1,2,2,3,4,4,3]
true

[1,2,2,null,3,null,3]
false

Note：

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

解釋下題目：

判斷一棵樹是不是鏡像的

1. 遞歸1

實際耗時：12ms

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

    public boolean isMirror(TreeNode left, TreeNode right) {
        if (left == null && right == null) {
            return true;
        } else if (left == null || right == null) {
            return false;
        } else {
            if (left.val == right.val) {
                return isMirror(left.left, right.right) && isMirror(left.right, right.left);
            } else {
                return false;
            }
        }
    }

踩過的坑：空樹

??思路很簡單，直接看就行

時間復雜度O(n)

空間復雜度O(n)