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  • 剑指 Offer 28. 对称的二叉树

    递归

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

    我的前方是万里征途,星辰大海!!
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  • 原文地址:https://www.cnblogs.com/taoyuxin/p/13471266.html
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