和上一题思路差不多，只不过有一个陷阱，就是不能用上一题的递归解法，因为逆序遍历会导致后面需要顺序的节点也逆序。

/*** 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>> result;vector<TreeNode*> tree;vector<vector<int>> zigzagLevelOrder(TreeNode* root) {if(root==NULL) return result;tree.push_back(root);while(!tree.empty()){int sum=tree.size();result.push_back(vector<int> ());for(int i=0;i<sum;i++){if(tree[i]->left) tree.push_back(tree[i]->left);if(tree[i]->right) tree.push_back(tree[i]->right);}if(result.size()%2!=1) for(int i=sum-1;i>=0;i--) result.back().push_back(tree[i]->val);else for(int i=0;i<sum;i++) result.back().push_back(tree[i]->val);tree.erase(tree.begin(),tree.begin()+sum);}return result;}
};