Binary Tree Level Order Traversal

단순히 트리를 BFS 방식으로 순회하여

레벨별로 노드의 값들을 벡터로 넣으면 되는 문제이다.

/**
 * 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>> levelOrder(TreeNode* root) {
        
        vector<vector<int>> result{};
        if (root == nullptr)
        {
            return result;
        }
        
        list<TreeNode *> treeList{root};
        
        while (treeList.empty() == false)
        {
            list<TreeNode *> levelTreeList{};
            swap(treeList, levelTreeList);
            
            vector<int> tempResult{};
            
            while (levelTreeList.empty() == false)
            {
                TreeNode *head = levelTreeList.front();
                levelTreeList.pop_front();
                
                tempResult.push_back(head->val);
                
                if (head->left != nullptr)
                {
                    treeList.push_back(head->left);
                }
                
                if (head->right != nullptr)
                {
                    treeList.push_back(head->right);
                }
                
            }
            
            result.push_back(tempResult);
            
        }
        
        return result;
    }
};