Leetcode 103. Binary Tree Zigzag Level Order Traversal（ C++版）

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]]

思路分析：

先按照正常的层序遍历的思想，然后把得到的结果的奇数层进行反转

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> > res;        if(!root) return res;                queue<TreeNode *> q;        q.push(root);        int level = 1;        while(!q.empty()){            vector<int> everylevel;            int size = q.size();            for(int i = 0; i < size; i ++){//不能直接q.size()                TreeNode * t = q.front();                everylevel.push_back(t -> val);                q.pop();                if(t -> left)                    q.push(t -> left);                if(t -> right)                    q.push(t -> right);            }            res.push_back(everylevel);        }        for(int i = 1; i < res.size(); i += 2){            reverse(res[i].begin(), res[i].end());        }        return res;    }};