**一. 题目描述** 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 ~~~ **二. 题目分析** BFS，这里用一个bool记录是从左到右还是从右到左读取数据，每一层遍历结束就翻转一下。 **三. 示例代码** ~~~ #include <iostream> #include <vector> using namespace std; 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> > result; Traverse(root, 1, result, true); return result; } private: void Traverse(TreeNode *root, size_t level, vector<vector<int> > & result, bool leftToRight) { if (root == NULL) return; if (level > result.size()) result.push_back(vector<int>()); if (leftToRight) result[level - 1].push_back(root->val); else result[level - 1].insert(result[level - 1].begin(), root->val); Traverse(root->left, level + 1, result, !leftToRight); Traverse(root->right, level + 1, result, !leftToRight); } }; ~~~ **四. 小结** 比Binary Tree Level Order Traversal 和Binary Tree Level Order Traversal II 稍微难一些，需要注意更多细节。