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 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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

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class Solution {
public:
    bool judge(TreeNode *p,TreeNode *q)
    {
        if(!p&&!q)return true;
        else  if(p&&q&&p->val==q->val) return judge(p->left,q->right)&&judge(p->right,q->left);
        else return false;
        
    }
    
    bool isSymmetric(TreeNode* root) {
        if(!root)return true;
        else return judge(root->left,root->right);
    }
};

非递归：<pre name="code" class="cpp">class Solution {
public:
    bool isSymmetric(TreeNode *root) {
       
        if (!root) return true;
        queue<TreeNode*> q1, q2;
        q1.push(root->left);
        q2.push(root->right);
        while (!q1.empty() && !q2.empty())
        {
            TreeNode *t1 = q1.front();
            TreeNode *t2 = q2.front();
            q1.pop();
            q2.pop();
            if (!t1 && !t2) continue;
            if (!t1|| !t2) return false;
            if(t1->val!=t2->val) return false;
            q1.push(t1->left);
            q1.push(t1->right);
            q2.push(t2->right);
            q2.push(t2->left);
        }
        return true;
    }
};