Leetcode-binary tree Zigzag level Order traversal two fork Tree Z-Scan

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]]

The topic is still relatively simple, mainly is a level scan and then the process of flipping

Here is a bit like a sequence scan, in the sequence scan is mainly using a queue for storage node, and then read out the corresponding Val, here needs to return the type needs to be given by layer, so you need to use a vector to each layer of nodes to record

/** * Definition for binary tree * struct TreeNode {* int val; * TreeNode *left; * TreeNode *right; * Tre Enode (int x): Val (x), left (null), right (NULL) {} *};    */class Solution {public:vector<vector<int>> Res;    int flag= 0;        vector<vector<int> > Zigzaglevelorder (TreeNode *root) {res.clear ();        if (root = NULL) {return res;        } vector<treenode*>pnode;        Pnode.push_back (root);        vector<int> tmp;        Tmp.push_back (Root->val);        Res.push_back (TMP);        flag++;        GetNode (Pnode);    return res;        } void GetNode (Vector<treenode*> &pnode) {if (Pnode.empty ()) {return;        } vector<treenode*> Node;        Vector<treenode*>::iterator it = Pnode.begin ();        vector<int> tmp;            for (; it! = Pnode.end (); it++) {TreeNode * root = *it;         if (root->left)   {Tmp.push_back (root->left->val);            Node.push_back (Root->left);                } if (root->right) {tmp.push_back (root->right->val);            Node.push_back (Root->right);            }} if (Tmp.empty ()) {return;//< No child node, can return} if (flag&1) {        Reverse (Tmp.begin (), Tmp.end ());        } flag++;        Res.push_back (TMP);    GetNode (Node); }};

The main use of a vector flip operation

Also see a way to operate with a queue, where a null node is placed on the last side of each layer of operations to determine that the data for this layer has been solved

/** * Definition for binary tree * struct TreeNode {* int val; * TreeNode *left; * TreeNode *right; * Tre Enode (int x): Val (x), left (null), right (NULL) {} *}; */class Solution {public:vector<vector<int> > Zigzaglevelorder (TreeNode *root) {VECTOR&LT;VECTOR&L        t;int> > result;        Vector<int> temp;        Result.clear ();        if (root==null) return result;        Queue<treenode *> ptemp;        int flag=0;        Ptemp.push (root);        Ptemp.push (NULL);            while (!ptemp.empty ()) {TreeNode *curnode=ptemp.front ();            Ptemp.pop ();                if (curnode!=null) {temp.push_back (curnode->val);                if (curnode->left) Ptemp.push (curnode->left);            if (curnode->right) Ptemp.push (curnode->right);          } else {if (!temp.empty ()) {          Ptemp.push (NULL);                    if (flag==1) {reverse (Temp.begin (), Temp.end ());                    } result.push_back (temp);                    Flag=1-flag;                Temp.clear ();    }}} return result; }};

Leetcode-binary tree Zigzag level Order traversal two fork Tree Z-Scan