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).
**Example :**Given binary tree
`3
/ \ 9 20 / \ 15 7`
return
[ [3], [20, 9], [15, 7] ]
ZigZag Level Order Traversal BT - InterviewBit
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
vector<vector<int> > Solution::zigzagLevelOrder(TreeNode* root) {
vector<vector<int>> res;
if(!root)
return res;
deque<TreeNode*> dq;
dq.push_front(root);
int turn = 0;
while(!dq.empty())
{
int lsize = dq.size();
vector<int> temp;
if(turn==0)
{
while(lsize--)
{
TreeNode *node = dq.front();
dq.pop_front();
temp.push_back(node->val);
if(node->left)
dq.push_back(node->left);
if(node->right)
dq.push_back(node->right);
}
}
else
{
while(lsize--)
{
TreeNode *node = dq.back();
dq.pop_back();
temp.push_back(node->val);
if(node->right)
dq.push_front(node->right);
if(node->left)
dq.push_front(node->left);
}
}
res.push_back(temp);
turn = 1-turn;
}
return res;
}