递归遍历
recursion(root1,root2){
if(root1 == null && root2 == null){return ture;}
if(root1 == null || root2 == null || root1.val != root2.val)return false;
returen recursion(root1.left,root2.right) && recursion(root2.left,root2.right);
}
标签：判断,return,recursion,二叉树,对称,null,root1,root2 From： https://www.cnblogs.com/materialdog/p/17301178.html