好久没刷题今天有时间挑选了这道Symmetric Tree(对称二叉树)。
按照最直观的思路层序遍历，然后判断每一层是否对称。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        queue<TreeNode *> q;
        if(root) q.push(root);
        while(!q.empty())
        {
            int n = q.size();
            vector<int> vec(n,0);
            for(int i = 0;i < n;i++)
            {
                TreeNode* node = q.front();
                q.pop();
                vec[i] = node->val;
                if(node->left) 
                    q.push(node->left);
                if(node->right) 
                    q.push(node->right);
            }
            if(!judgeSymmetric(vec)) return false;
        }
        return true;
    }
private:
    bool judgeSymmetric(const vector<int>& vec)
    {
        int n = vec.size();
        for(int i = 0;i < n / 2;i++)
        {
            if(vec[i] != vec[n-1-i]) return false;
        }
        return true;
    }
};

最后发现有一种case无法处理，形如[1,2,2,null,3,null,3]等等情况。
正确的解法是使用层序遍历的变种：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        return isSymmetric(root,root);
    }
private:
    bool isSymmetric(TreeNode* left,TreeNode* right) {
        queue<TreeNode*> q;
        q.push(left),q.push(right);
        while(!q.empty())
        {
            TreeNode* u = q.front();q.pop();
            TreeNode* v = q.front();q.pop();
            //都是空指针,继续处理
            if(!u && !v) continue;
            //1.其中一个指针为空 或 值不相等返回false
            if((!u || !v) || u->val != v->val) return false;
            //插入顺序:left：从左向右，right:从右向左
            q.push(u->left);q.push(v->right);
            q.push(u->right);q.push(v->left);
        }
        return true;
    }
};

递归实现：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        return isSymmetric(root,root);
    }
private:
    bool isSymmetric(TreeNode* p,TreeNode* q) {
        if(!p && !q) return true;
        if(!p || !q) return false;
        
        return (p->val == q->val) && isSymmetric(p->left,q->right) && isSymmetric(p->right,q->left);
    }
};