题解 | #31.对称的二叉树#

递归

条件：p1.val == p2.val && compare(p1.left,p2.right) && compare(p1.right,p2.left)

function isSymmetrical(pRoot)
{
  function compare(p1,p2){
    if(p1==null && p2==null)
      return true;
    if(p1==null || p2==null)
      return false;
    return p1.val == p2.val && compare(p1.left,p2.right) && compare(p1.right,p2.left);
  }
  if(pRoot == null)
    return true;
  return compare(pRoot.left,pRoot.right);
}
module.exports = {
    isSymmetrical : isSymmetrical
};