[Leetcode] Binary Tree Zigzag level Order traversal

"title"

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    /   7

Return its zigzag level order traversal as:

[  3],  [20,9],  [15,7]]

Confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

"Code One"

/********************************** Date: 2014-12-09* sjf0115* question number: Binary Tree Zigzag level Order traversal* Source: HTTPS ://oj.leetcode.com/problems/binary-tree-zigzag-level-order-traversal/* Results: ac* Source: leetcode* Summary: ******************* /#include <iostream> #include <malloc.h> #include <stack> #include <vector>#    Include <queue>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> > Zigzagle        Velorder (TreeNode *root) {vector<int> level;        Vector<vector<int> > levels;        if (root = NULL) {return levels;        } queue<treenode*> CURQ,NEXTQ;        Stack<treenode*> curs,nexts;        Index layer int index = 1;        Into the queue curq.push (root);     Hierarchical traversal while (!curq.empty () | | |!curs.empty ()) {//Current layer traversal       while (!curq.empty () | |!curs.empty ()) {TreeNode *p,*q;                    The odd layer uses the queue to store if (index & 1) {p = Curq.front ();                    Curq.pop (); The first even tier stores the next layer of nodes with stacks//left dial hand tree if (p->left) {Nexts.push (P->left)                        ;                    Used to traverse the node Nextq.push (P->left) from left to right;                        }//Right subtree if (p->right) {Nexts.push (p->right);                    Used to traverse Nextq.push (P->right) from left to right;                    }}//Even layer with stack storage else{p = curs.top ();                    Curs.pop ();                    The storage node is traversed from left to right through q = Curq.front ();                    Curq.pop ();  The odd layer uses the queue to store the next layer of nodes//left dial hand tree if (q->left) {                      Nextq.push (Q->left);                    }//Right subtree if (q->right) {Nextq.push (q->right);            }} level.push_back (P->val);            } index++;            Levels.push_back (level);            Level.clear ();            Swap (NEXTQ,CURQ);        Swap (nexts,curs);    }//while return levels;    }};//creates a two-fork-tree int createbtree (treenode* &t) {char data) in order sequence;    Enter the value of the node in the binary tree in order of precedence (one character), and ' # ' indicates the empty tree cin>>data;    if (data = = ' # ') {T = NULL;        } else{T = (treenode*) malloc (sizeof (TreeNode));        Generate root node T->val = data-' 0 ';        Constructs left subtree Createbtree (t->left);    Construct right subtree Createbtree (t->right); } return 0;}    int main () {solution solution;    treenode* root (0);    Createbtree (root);    vector<vector<int> > VECs = solution.zigzaglevelorder (root); for (int i = 0;i< Vecs.size (); i++) {for (int j = 0;j < Vecs[i].size (); j + +) {cout<<vecs[i][j];    } cout<<endl; }}

"code two"

[Leetcode] Binary Tree Zigzag level Order traversal