binary_tree_level_order_traversal.cpp

/*
102. Binary Tree Level Order Traversal
Medium

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

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

[
  [3],
  [9,20],
  [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) {}
 * };
 */

//We use BFS using a queue
//We have a for loop that runs upto size (number of nodes in the queue)
//(this represents how many nodes are on the current level of the tree)
//After adding their value to the list, check if they have left and right children 
//adding them to the queue if they do exist 
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> result;
        if(root==NULL)
            return result;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()){
            vector<int> temp;
            int size = q.size();
            for(int i=0;i<size;i++){
                TreeNode* current = q.front();
                q.pop();
                temp.push_back(current->val);
                if(current->left!=NULL){
                    q.push(current->left);
                }
                if(current->right!=NULL){
                    q.push(current->right);
                }
            }
            result.push_back(temp);
        }
        return result;
            
    }
};