101. Symmetric Tree--judging whether the structure of trees is symmetrical

Given a binary tree, check whether it is a mirror of the itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

    1   /   2   2/\/3  4 4  3

But the following are not:

    1   /   2   2   \      3    3

Note:
Bonus points if you could solve it both recursively and iteratively.

1. Recursion

bool issymmetric (TreeNode *p, TreeNode *q) {    ifreturntrue;     if return false ;         return (P->val = = q->val) &&            issymmetric (P->left, q->right) &&             Issymmetric (P->right, q-> left);}

2. Non-recursive

BOOLIssymmetric (TreeNode *p, TreeNode *P) {Queue<TreeNode*>Q1; Queue<TreeNode*>Q2;    Q1.push (P);    Q2.push (q);  while(Q1.size () >0&& q2.size () >0) {TreeNode* P1 =Q1.front ();        Q1.pop (); TreeNode* P2 =Q2.front ();        Q2.pop (); if(P1==null && P2==null)Continue; if(P1==null | | p2==null)return false; if(P1->val! = p2->val)return false; Q1.push (P1-Left ); Q2.push (P2-Right ); Q1.push (P1-Right ); Q2.push (P2-Left ); }    return true;}

101. Symmetric Tree--judging whether the structure of trees is symmetrical