对称二叉树

WA1

测试案例：192/193通过率。。。。

/**
 * 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:
    std::vector<int>inorder1;
    std::vector<int>inorder2;
    bool isSymmetric(TreeNode* root) {
        if(root==NULL)return true;
        TreeNode* root2 = root;
        in1(root);
        in2(root2);
        bool ans = true;
        for(int i = 0;i<inorder1.size();i++){
            if(inorder1[i]!=inorder2[i]){
                ans = false;
                break;
            }
        }
    return ans;
}
    void in1(TreeNode *root) {
        if (root->left)
            in1(root->left);
        else
            inorder1.push_back(-INT_MAX);
        inorder1.push_back(root->val);
        if (root->right)
            in1(root->right);
        else
            inorder1.push_back(INT_MAX);
    }
    void in2(TreeNode *root) {
        if (root->right)
            in2(root->right);
        else
            inorder2.push_back(-INT_MAX);
        inorder2.push_back(root->val);
        if (root->left)
            in2(root->left);
        else
            inorder2.push_back(INT_MAX);
    }
};

WA2

122 / 193 个通过测试用例

bool isSymmetric(TreeNode *root) {
    if(root==NULL)
        return true;
    queue<TreeNode*> q;
    q.push(root);
    while(!q.empty()){
        int size = q.size();
        vector<int> subqueue;
        for (int i = 0; i < size; i++) {
            TreeNode* node = q.front();
            subqueue.push_back(node->val);
            q.pop();
            if(node->left)
                q.push(node->left);
            else
                subqueue.push_back(INT_MAX);
            if(node->right)
                q.push(node->right);
            else
                subqueue.push_back(-INT_MAX);
        }
        vector<int> subqueue2(subqueue);
        reverse(subqueue2.begin(),subqueue2.end());
        for (int j = 0; j < subqueue.size(); ++j) {
            if(subqueue[j]!=subqueue2[j])
                return false;
        }
    }
    return true;

AC1

    bool isSymmetric(TreeNode* root) {
       return isSymmetric2(root,root);
    }
    bool isSymmetric2(TreeNode* root1,TreeNode* root2){
      if(root1==NULL&&root2==NULL)return true;
      if(root2==NULL||root1==NULL)return false;
      return (root1->val==root2->val)&&isSymmetric2(root1->left,root2->right)&&isSymmetric2(root1->right,root2->left);
    }