101. Symmetric Tree -- 判断树结构是否对称

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

For example, this binary tree is symmetric:

    1
   / 
  2   2
 /  / 
3  4 4  3

But the following is not:

    1
   / 
  2   2
      
   3    3

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

1. 递归

bool isSymmetric(TreeNode *p, TreeNode *q){
    if (p==NULL && q==NULL) return true;
    if (p==NULL || q==NULL) return false;
    
    return (p->val == q->val) &&
            isSymmetric(p->left, q->right) &&
            isSymmetric(p->right, q->left);
}

2. 非递归

bool isSymmetric(TreeNode *p, TreeNode *q)
{
    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;
}