leetcode-Binary Tree Zigzag Level Order Traversal

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.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1  / \ 2   3    /   4    \     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

基础题，二叉树的层次遍历。

/** * Definition for binary tree * 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) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        vector<vector<int> > ret;                if (root == NULL)            return ret;                vector<int> nodes;                queue<TreeNode*> q;        TreeNode *nodePtr;        // levelNodesNum: the number of nodes in next level to traverse        int levelNodesNum = 1, nums, i;              int zz = 0;   // If 1, from left to right, or, from right to left                q.push(root);        nodes.push_back(root->val);        ret.push_back(nodes);                while (!q.empty()) {            i = 0;            nums = 0;            nodes.clear();                        // Traverse the nodes of the same level, and push the nodes of next level            while (i < levelNodesNum) {                   nodePtr = q.front();                if (nodePtr->left != NULL) {                    q.push(nodePtr->left);                    nodes.push_back(nodePtr->left->val);                    ++nums;                }                if (nodePtr->right != NULL) {                    q.push(nodePtr->right);                    nodes.push_back(nodePtr->right->val);                    ++nums;                }                q.pop();                ++i;            }                        levelNodesNum = nums;  // Update the number of nodes of the next level                        if (nodes.empty()) {                break;            }                        if (zz) {                   ret.push_back(nodes);                     zz = 0;            } else {                reverse(nodes.begin(), nodes.end());   // Right to left                ret.push_back(nodes);                zz = 1;            }        }                return ret;    }};