问题描述：
其实就是给了一个二叉搜索树，然后按照龙摆尾的方式输出出来，在层序遍历的方式进行改变，其实就增加了两行代码，一个是flag标志位，一个是按照flag的true与false来更改存储顺序。
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] ]
C++
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Solution {
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public:
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vector<vector<int> > zigzagLevelOrder(TreeNode* root)
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{
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if(root==NULL)
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return vector<vector<int> > ();
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vector<vector<int>>res;
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//用队列存储这个层的结点
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queue<TreeNode*>queue;
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queue.push(root);
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//作为奇偶行的判断标志
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bool flag=true;
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while(!queue.empty())
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{
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int size=queue.size();//该层有多少结点
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vector<int>row(size);
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for(int i=0;i<size;i++)
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{
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TreeNode*tempnode=queue.front();
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queue.pop();
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if(tempnode->left)
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{
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queue.push(tempnode->left);
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}
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if(tempnode->right)
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{
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queue.push(tempnode->right);
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}
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//把该结点按照龙摆尾的条件存储
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//方法就是在下表做文章，从最高位开始存储，然后下一次再从低位开始
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int idex=(flag)?i:size-1-i;
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row[idex]=tempnode->val;
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}
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//该层已经能够循环完了 接下来就是下一蹭了
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//把改行数据存起来,千万不要忘记更改flag
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flag=!flag;
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res.push_back(row);
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}
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return res;
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}
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};
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/* if (root == NULL) {
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return vector<vector<int> > ();
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}
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vector<vector<int> > result;
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queue<TreeNode*> nodesQueue;
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nodesQueue.push(root);
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bool leftToRight = true;
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while ( !nodesQueue.empty()) {
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int size = nodesQueue.size();
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vector<int> row(size);
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for (int i = 0; i < size; i++) {
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TreeNode* node = nodesQueue.front();
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nodesQueue.pop();
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// find position to fill node's value
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int index = (leftToRight) ? i : (size - 1 - i);
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row[index] = node->val;
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if (node->left) {
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nodesQueue.push(node->left);
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}
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if (node->right) {
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nodesQueue.push(node->right);
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}
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}
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// after this level
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leftToRight = !leftToRight;
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result.push_back(row);
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}
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return result;
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}*/