69. Binary Tree Level Order Traversal

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param root: A Tree
     * @return: Level order a list of lists of integer
     */
    vector<vector<int>> levelOrder(TreeNode * root) {
        vector<vector<int>> result;
        if(root == NULL){
            return result;
        }
        std::queue<TreeNode *>  Q;
        Q.push(root);
        while(!Q.empty()){
            vector<int> level;
            int size = Q.size();
            std::cout <<"size " << size << " " << "Q.size() " << Q.size() << std::endl;
            for (int i = 0; i < size; i++) {
                
               TreeNode *head = Q.front();
                Q.pop();
                level.push_back(head->val);
                if(head->left != NULL){
                    Q.push(head->left);
                }
                if(head->right != NULL){
                    Q.push(head->right);
                }
            }
            result.push_back(level);
        }
        return result;
    }
};

Python

from collections import deque
"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""

class Solution:
  
    def levelOrder(self, root):
        if root is None:
            return []

        queue = deque([root])
        result = []
        while queue:
            level = []
            for _ in range(len(queue)):
                node = queue.popleft()
                level.append(node.val)
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            result.append(level)
        return result