力扣精选top面试题---------二叉树的锯齿形层序遍历

题目链接！

思路：
这道题我们可以用队列来做的，你会发现只要我们每两层之间翻转一层即可，基本的框架还是bfs，具体可以看代码。

代码：

/**
 * 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:
    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
    
    
        vector<vector<int>> ans;
        if (!root) {
    
    
            return ans;
        }

        queue<TreeNode*> nodeQueue;
        nodeQueue.push(root);
        bool isOrderLeft = true;

        while (!nodeQueue.empty()) {
    
    
            vector<int> levelList;
            int size = nodeQueue.size();
            for (int i = 0; i < size; ++i) {
    
    
                auto node = nodeQueue.front();
                nodeQueue.pop();
                levelList.push_back(node->val);

                if (node->left) {
    
    
                    nodeQueue.push(node->left);
                }
                if (node->right) {
    
    
                    nodeQueue.push(node->right);
                }
            }
            if(isOrderLeft)
                ans.emplace_back(vector<int>{
    
    levelList.begin(), levelList.end()});
            else
                ans.emplace_back(vector<int>{
    
    levelList.rbegin(), levelList.rend()});
            isOrderLeft = !isOrderLeft;
        }

        return ans;
    }
};