LeetCode 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 [1,2,2,3,4,4,3] is symmetric:

1
/ \
2   2
/ \ / \
3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

1
/ \
2   2
\   \
3    3

Note:

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

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isSymmetric(TreeNode* root) {
vector<int>a;
vector<int>b;
if(root==NULL)return true;
else{
if(root->left!=NULL&&root->right!=NULL){
a=solvel(root->left,a);
b=solver(root->right,b);

}
else if(root->left==NULL&&root->right==NULL)return true;
else return false;
}
return a==b;
}
vector<int> solvel(TreeNode *root,vector<int>r){
r.push_back(root->val);
if(root->left!=NULL||root->right!=NULL){

if(root->left!=NULL&&root->right==NULL){r.push_back(NULL);r=solvel(root->left,r);}
if(root->left==NULL&&root->right!=NULL){r=solvel(root->right,r);r.push_back(NULL);}
if(root->left!=NULL&&root->right!=NULL){
r=solvel(root->left,r);
r=solvel(root->right,r);

}
}

return r;
}
vector<int> solver(TreeNode *root,vector<int>r){
r.push_back(root->val);
if(root->left!=NULL||root->right!=NULL){

if(root->left!=NULL&&root->right==NULL){r=solver(root->left,r);r.push_back(NULL);}
if(root->left==NULL&&root->right!=NULL){r.push_back(NULL);r=solver(root->right,r);}
if(root->left!=NULL&&root->right!=NULL){
r=solver(root->right,r);
r=solver(root->left,r);
}
}

return r;
}
};

这是大神们写的：

class Solution {
public:
bool isSymmetric(TreeNode *root) {
TreeNode *left, *right;
if (!root)
return true;

queue<TreeNode*> q1, q2;
q1.push(root->left);
q2.push(root->right);
while (!q1.empty() && !q2.empty()){
left = q1.front();
q1.pop();
right = q2.front();
q2.pop();
if (NULL == left && NULL == right)
continue;
if (NULL == left || NULL == right)
return false;
if (left->val != right->val)
return false;
q1.push(left->left);
q1.push(left->right);
q2.push(right->right);
q2.push(right->left);
}
return true;
}
};

大神的思路是每次比较一个元素，我的这个是全存起来再比较。