leetcode 101 Symmetric Tree C++

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递归版

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

非递归版

class Solution {public:    bool isSymmetric(TreeNode* root) {        queue<TreeNode *> que;        vector<vector<int>> vec;        if (!root) return true;        que.push(root);        while(que.size() != 0){            vector<int> tmpVec;            queue<TreeNode*> tmpQue;            while(que.size() != 0){                TreeNode * node = que.front();                que.pop();                                if(node->left) tmpQue.push(node->left);                else if(!node->left && node->val != -100) tmpQue.push(new TreeNode(-100));                if(node->right) tmpQue.push(node->right);                else if(!node->right && node->val != -100) tmpQue.push(new TreeNode(-100));                                tmpVec.push_back(node->val);            }            vec.push_back(tmpVec);            que = tmpQue;        }                        for (int i = 1;i<vec.size();i++){            vector<int> tmp = vec[i];            if(tmp.size()%2 != 0) return false;            for(int j = 0;j<tmp.size();j++){                if(tmp[j] == tmp[tmp.size()-j-1]) continue;                else return false;            }        }        return true;    }};

这个非递归版是可以骗过机器，但是可能会出问题，选择的值应该是不会出现的值才行。而且效率也不高。。。

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