這道題我給出兩種解法：
第一種，是使用分層儲存二叉樹，然後奇數層不翻轉，偶數層翻轉

/**
 * 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;
        res = levelOrder(root);

        for(int i = 0; i < res.size(); ++i)
        {
            if(i & 0x1 != 0)
            {
                reverse(res[i].begin(), res[i].end());
            }
        }
        return
 
 res;
    }
    vector<vector<int> > levelOrder(TreeNode* root)
    {
        int nextlevel = 0;
        int tobeprinted = 1;
        vector<int> curlevel;
        vector<vector<int> > res;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty())
        {
            curlevel.push_back(q.front() -> val);
            if
 
(q.front() -> left != NULL)
            {
                q.push(q.front() -> left);
                nextlevel++;

            }
            if(q.front() -> right != NULL)
            {
                q.push(q.front() -> right);
                nextlevel++;
            }
            q.pop();
            tobeprinted--;
            if(tobeprinted == 0)
            {
                res.push_back(curlevel);
                curlevel.clear();
                tobeprinted = nextlevel;
                nextlevel = 0;
            }

        }
        return res;

    }
};

還有一種是劍指offer上利用兩個棧的解法

class Solution {
public:
    vector<vector<int> > zigzagLevelOrder(TreeNode* root) {
        stack<TreeNode*> s1;
        stack<TreeNode*> s2;
        vector<int> curlevel; 
        vector<vector<int> > res;
        int tobeprinted = 1;   //tobeprinted 和 nextlevel 用於標記換行
        int nextlevel = 0;
        int levelnum = 1; //用於指示當前的行號，從1開始
        if(root == NULL)
            return res;
        s1.push(root);
        while(!s1.empty() || !s2.empty())
        {
            if((levelnum & 0x1) == 1)
            {
                if(!s1.empty())
                {
                    if(s1.top() != NULL)
                    {
                        curlevel.push_back(s1.top() -> val);
                        if(s1.top() -> left != NULL)
                        {
                            s2.push(s1.top() -> left);
                            nextlevel++;
                        }

                        if(s1.top() -> right != NULL)
                        {
                            s2.push(s1.top() -> right);
                            nextlevel++;
                        }
                        s1.pop();

                    }

                }
            }
            else
            {
                if(!s2.empty())
                {
                    if(s2.top() != NULL)
                    {
                        curlevel.push_back(s2.top() -> val);
                        if(s2.top() -> right != NULL)
                        {
                            s1.push(s2.top() -> right);
                            nextlevel++;
                        }

                        if(s2.top() -> left != NULL)
                        {
                            s1.push(s2.top() -> left);
                            nextlevel++;
                        }
                        s2.pop(); 

                    }


                }
            }
            tobeprinted--;
            if(tobeprinted == 0)
            {
                res.push_back(curlevel);
                curlevel.clear();
                tobeprinted = nextlevel;
                nextlevel = 0;
                levelnum++;
            }
        }
        return res;

    }
};