Day-22-Binary Tree Zigzag Level Order Traversal.cpp

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]
]






/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
        vector<vector<int>> res;
        if(root== NULL)
            return res;
        queue<TreeNode*> q;
        q.push(root);
        int f = false;
        while(q.size())
        {
            int size = q.size();
            vector<int> level;
            while(size--)
            {
                TreeNode* curr = q.front();
                level.push_back(curr->val);
                q.pop();
                if(curr->left)
                {
                    q.push(curr->left);
                }
                if(curr->right)
                {
                    q.push(curr->right);
                }

            }
            if(f)
             reverse(level.begin(),level.end());
            f=!f;
            res.push_back(level);
        }
        return res;
    }
};