103. 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,null,null,15,7],

    3   / \  9  20    /  \   15   7

return its zigzag level order traversal as:

[  [3],  [20,9],  [15,7]]

折线遍历树，那我们可以对树进行层次遍历。定义一个flag，表示是否是从左往右打印，如果不是，reverse即可。第一次循环，打印root，flag=true，无需reverse。flag取反。第二次循环，应该从左往右，所以会reverse，迭代下去即可。

这个算法的亮点是利用队列que.size()控制树每层元素的个数，让我们可以知道push的次数。保证每个分支都能入队。

代码如下：

/** * 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>> res;        if(root == NULL)            return res;                queue<TreeNode*> que;        que.push(root);        bool LR = true;                while(!que.empty()){            vector<int> vec;            int size = que.size();                 for(int i=0; i<size; ++i){                TreeNode* t = que.front();                que.pop();                            vec.push_back(t->val);                            if(t->left != NULL)                    que.push(t->left);                if(t->right != NULL)                    que.push(t->right);            }                        if(!LR)                reverse(vec.begin(), vec.end());            res.push_back(vec);            LR = !LR;        }        return res;    }};

PS：终于把缺的补回来了。