Leetcode-101: Symmetric Tree

This question is also a classic question and needs attention. My solution also refers to the answers on the Internet.
Solution 1: Recursion

bool isMirror(TreeNode* r1, TreeNode* r2) {
    if (!r1 && !r2) return true;
    if (!r1 || !r2) return false;
    if (r1->val!=r2->val) return false;
    return isMirror(r1->left, r2->right) && isMirror(r1->right, r2->left);

}

bool isSymmetric(TreeNode* root) {

    if (!root) return true;
    return isMirror(root->left, root->right);
}

Note: You can also use if() in isSymmetric(), but directly return isMirror(root, root). In leetcode, the version without if only takes 8 seconds, and my version takes 9 seconds. I don’t know why it is missing once Is recursion slower?

Solution 2: Iterate. It is similar to traversing the binary tree level.

bool isSymmetric(TreeNode* root) {
    queue<TreeNode*> q;
    if (!root) return true;
    q.push(root->left);
    q.push(root->right);
    while(!q.empty()) {
        TreeNode* p1=q.front(); q.pop();
        TreeNode* p2=q.front(); q.pop();
        if (!p1 && !p2) continue;  //注意这里要返回。
        if (!p1 || !p2) return false;
        if (p1->val != p2->val) return false;
        q.push(p1->left);
        q.push(p2->right);
        q.push(p1->right);
        q.push(p2->left);
    }

    return true;
}

Note: Some solutions on the Internet here do not check if(!root) at first, but push(root) twice. I tried it, this version takes 10 seconds on leetcode, my version above is only push(root) once, and it takes 8 seconds.