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   7return its zigzag level order traversal as:[  [3],  [20,9],  [15,7]]遍历二叉树，按照奇数行从左向右，偶数行从右向左，之字形遍历。如何判断何时什么时候从左向右，什么时候从右向左，需要额外的记录该信息；二叉树按行遍历，BFS，可以用队列实现，队列中每次都读取当前行时，存储下一行，每一行开始读取时，取出现在队列的大小即该行有多少个节点，而节点读取后存储的顺序为index = reverse ? n-i-1 : i，index为存储的位置，n为该行的大小,i为在队列中的位置，reverse为是否反序。*//** * 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;        queue<TreeNode*> q;        bool reverse = false;        if(root) q.push(root);        while(!q.empty())        {            int n = q.size();            vector<int> tmp(n);            for(int i=0;i<n;i++)            {                int index = reverse ? n-i-1 : i;                tmp[index]=q.front()->val;                if(q.front()->left) q.push(q.front()->left);                if(q.front()->right) q.push(q.front()->right);                q.pop();            }            res.push_back(tmp);            reverse=!reverse;        }        return res;    }};