101. Symmetric binary tree

Punch in one question every day !!!
Today I want to share with you a topic of recursive type .

Title Description ：

Give you the root node of a binary tree root , Check whether it is axisymmetric .

Title Example ：

Whether it is axisymmetric , In fact, let's judge whether the subtrees on the left and right sides are equal .
In a little bit more depth , In fact, let us judge

• left The left node of and right Whether the right nodes of are equal
• left The right node of and right Whether the left nodes of are equal .

I use the example in the title 1 For example , On the left side of the second floor 2 This node , Its left node =3;
On the right side of the second floor 2 This node , Its right node =3; At present, it still conforms to symmetry , Let's continue to judge downward ：
On the left side of the second floor 2 This node , Its right node =4;
On the right side of the second floor 2 This node , Its left node =4;
All meet the requirements of symmetry , It means that the third layer is symmetrical , If there is a third layer 、 The fourth level 、 The fifth floor … We can continue to judge downward according to this logic .

Code ：(PS：Java edition )

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

    }

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