101. Symmetric Tree

101. Symmetric Tree
     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
python

def isSymmetric(self, root):
    if not root:
        return True
    return self.dfs(root.left, root.right)
    
def dfs(self, l, r):
    if l and r:
        return l.val == r.val and self.dfs(l.left, r.right) and self.dfs(l.right, r.left)
    return l == r

java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root ==  null)
            return true;
        return dfs(root.left,root.right);
    }
    private boolean dfs(TreeNode l,TreeNode r){
        if(l != null && r!= null){
            return dfs(l.left,r.right) && dfs(l.right,r.left) && l.val == r.val;
        }
        return l == r;
    }
}