Q26_binary_tree_level_order_traversal.cpp

// Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level).

// Example 1:
// Input: root = [3,9,20,null,null,15,7]
// Output: [[3],[9,20],[15,7]]

// Example 2:
// Input: root = [1]
// Output: [[1]]

// Example 3:
// Input: root = []
// Output: []

#include<bits/stdc++.h>
using namespace std;

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>> levelOrder(TreeNode* root) {
        vector <vector<int>> v;
        if(root == NULL)
            return v;
        queue <TreeNode*> q;
        q.push(root);
        while(!q.empty()){
            int count = q.size();
            vector <int> temp;
            for(int i=0; i<count; i++){
                TreeNode *curr = q.front();
                q.pop();
                temp.push_back(curr->val);
                if(curr->left != NULL)
                    q.push(curr->left);
                if(curr->right != NULL)
                    q.push(curr->right);
            }
            v.push_back(temp);
        }
        return v;
    }
};

// Time Complexity  : O(n) 
// Space Complexity : O(w)
// where w is the width of the tree