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).
/**
* 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<TreeNode *>> levels;
vector<TreeNode *> parent_vector;
vector<TreeNode *> children_vector;
if(root)parent_vector.push_back(root);
while(parent_vector.size()>0){
levels.push_back(parent_vector);
children_vector.clear();
while(parent_vector.size()>0){
if(parent_vector[0]->left) children_vector.push_back(parent_vector[0]->left);
if(parent_vector[0]->right) children_vector.push_back(parent_vector[0]->right);
parent_vector.erase(parent_vector.begin());
}
parent_vector = children_vector;
}
vector<vector<int>> levels_vals;
for(int i = 0; i<levels.size(); i++){
vector<int> current_level_vals;
for(int j = 0; j<levels[i].size(); j++){
current_level_vals.push_back(levels[i][j]->val);
}
levels_vals.push_back(current_level_vals);
}
return levels_vals;
}
};