LeetCode#103. Binary Tree Zigzag Level Order Traversal My Submissions Question

103. Binary Tree Zigzag Level Order Traversal My Submissions Question
Total Accepted: 57077 Total Submissions: 201598 Difficulty: Medium
Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).


For example:
Given binary tree {3,9,20,#,#,15,7},
    3
   / \
  9  20
    /  \
   15   7
return its zigzag level order traversal as:
[
  [3],
  [20,9],
  [15,7]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

c++代码：

/** * 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:    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {        vector<vector<int>> Ans;//存放最后结果        if(root==NULL)return Ans;//根节点为空        queue<TreeNode*> Q;        Q.push(root);//根节点进队列        int countLayer=0;//标志 是否需要反转 奇数：需要 偶数：不需要        while(!Q.empty()){            queue<TreeNode*> tempQ;//存放下一层的结点队列（缓存）            vector<int> temp;//存放当前层的值            while(!Q.empty()){                TreeNode *head=Q.front();//取队首元素                Q.pop();                temp.push_back(head->val);                if(head->left!=NULL)tempQ.push(head->left);//左孩子非空，进入下一层的结点队列（缓存）                if(head->right!=NULL)tempQ.push(head->right);//右孩子非空，进入下一层的结点队列（缓存）            }            if(countLayer%2!=0)reverse(temp.begin(),temp.end());//需要反转            Ans.push_back(temp);//保存一层结点的值            ++countLayer;            Q=tempQ;//遍历下一层        }        return Ans;    }};