102. Binary Tree Level Order Traversal

1. Description

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

2. Solutions

My Accepted Solution

n is the number of nodes in i_root
Time complexity: O(n)
Space complexity: O(n)

class Solution 
{
public:
    // vector<vector<int>> levelOrder(TreeNode* root)
    vector<vector<int>> levelOrder(const TreeNode *i_root) 
    {
        if(!i_root) return vector<vector<int>>{};
        
        vector<vector<int>> result; 
        queue<const TreeNode *> currentLevelNodes{{i_root}};
        queue<const TreeNode *> nextLevelNodes;
        
        vector<int> levelNodesValue;
        while(!currentLevelNodes.empty() || !nextLevelNodes.empty())
        {
            if(currentLevelNodes.empty())
            {
                result.push_back(levelNodesValue);
                levelNodesValue.clear();
                
                swap(currentLevelNodes, nextLevelNodes);  
            }
            else
            {
                auto node = currentLevelNodes.front();
                currentLevelNodes.pop();
                
                levelNodesValue.push_back(node->val);
                
                if(node->left) nextLevelNodes.push(node->left);
                if(node->right) nextLevelNodes.push(node->right);
            }
        }
        
        result.push_back(levelNodesValue);
        
        return result;
    }
};
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