Leetcode-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  3But the following [1,2,2,null,3,null,3] is not:    1   / \  2   2   \   \   3    3

/** * 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:    void mirror(TreeNode* root){        if(root==NULL) return;        swap(root->left,root->right);        mirror(root->left);        mirror(root->right);    }    void preOrder(TreeNode* root,vector<int> &result){        if(root==NULL) return;        result.push_back(root->val);        preOrder(root->left,result);        preOrder(root->right,result);    }    void inOrder(TreeNode* root,vector<int> &result){        if(root==NULL) return;        inOrder(root->left,result);        result.push_back(root->val);        inOrder(root->right,result);    }    bool equalVector(vector<int> first, vector<int> second){        if(first.size()!=second.size()) return false;        for(int i=0;i<first.size();i++){            if(first[i]!=second[i])            return false;        }        return true;    }    bool equalTree(TreeNode* left, TreeNode* right){        vector<int> l_pre,l_in,r_pre,r_in;        preOrder(left,l_pre);        inOrder(left,l_in);        preOrder(right,r_pre);        inOrder(right,r_in);        return equalVector(l_pre,r_pre)&&equalVector(l_in,r_in);    }    bool isSymmetric(TreeNode* root) {        if(root==NULL) return true;        mirror(root->right);        return equalTree(root->left,root->right);    }};

以上方法虽然通过，但是有bug
bug用例
[1,1]
[1,null,1]

错误在与我判断两棵树是否相同的时候，用了 “两棵树的preorder & inorder 一致的话，则两棵树相等”，实际上有问题的，当节点的值相等，则此法无效。

正确做法如下：

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