题解 | #按之字形顺序打印二叉树#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
#include <algorithm>
#include <queue>
#include <vector>
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param pRoot TreeNode类 
     * @return int整型vector<vector<>>
     */

    vector<vector<int>> visit_seq(TreeNode* root)
    {
        vector<TreeNode*> v;
        v.push_back(nullptr);
        v.push_back(root);
        
        queue<TreeNode*> q;
        q.push(root);
        while (!q.empty()) {    // 层序遍历，把节点都装进容器里面，而且每层使用空节点间开
            v.push_back(nullptr);
            // std::cout << " v size = " << v.size() << std::endl;
            int n = q.size();
            // std::cout << " n = " << n << std::endl;
            for(int i = 0; i < n; ++i)
            {
                TreeNode* tmp = q.front();
                q.pop();
                if(tmp->left){
                    v.push_back(tmp->left);
                    // std::cout << tmp->left->val << " ";
                    q.push(tmp->left);
                }
                if(tmp->right){
                    v.push_back(tmp->right);
                    // std::cout << tmp->right->val << " ";
                    q.push(tmp->right);
                }
            }
            // std::cout << std::endl<< " *********** " << std::endl;
        }
        v.pop_back();
        // std::cout << v.size() << std::endl;
        vector<int> res_1;  // 存储同一层的节点的val值
        vector<vector<int>> res;
        int n = v.size();
        bool flag = true; // 从左到右
        for(int i = 0; i < n; ++i)
        {
            if (v[i]) {     // 表示同一层节点
                res_1.push_back(v[i]->val);
            }
            if(!v[i] && !res_1.empty()) // 表示到达下一层节点，将上一层节点的val存入容器中，使用flag来记录遍历顺序
            {
                // std::cout << std::endl << " *** " << std::endl;
                flag = !flag;
                if(flag) reverse(res_1.begin(), res_1.end());   // 
                res.push_back(res_1);
                res_1.clear();
            }
            if(i == n-1)
            {
                flag = !flag;
                if(flag) reverse(res_1.begin(), res_1.end());
                res.push_back(res_1);
                res_1.clear();
            }
        }
        return res;
    }

    vector<vector<int>> Print(TreeNode* pRoot) {
        // write code here
        TreeNode* root = pRoot;
        if(!root) return {};
        return visit_seq(root);

    }
};