102 - Binary Tree Level Order Traversal

#medium

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).

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: [ [[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) {}
 * };
 */
class Solution {
public:
    int getHeight( TreeNode * root ) {
        if ( root == NULL ) return 0;
        int height_left = getHeight( root -> left );
        int height_right = getHeight( root -> right );
        return max( height_left, height_right ) + 1;
    }

    // int tr_height = getHeight(root);
    // int max_nodes = pow( 2, tr_height ) - 1;

    vector<vector<int>> levelOrder(TreeNode* root) {
        queue<TreeNode*> que;
        if ( root != NULL ) que.push(root);
        vector<vector<int>> vv_res;
        while ( !que.empty() ) {
            int que_size = que.size();
            vector<int> one_level_value = {};
            for ( int i = 0; i < que_size; i++ ) {
                TreeNode * node = que.front();
                que.pop();
                one_level_value.push_back( node -> val );
                if ( node -> left != NULL ) que.push( node -> left );
                if ( node -> right != NULL ) que.push( node -> right );
            }

            vv_res.push_back(one_level_value);
        }

        return vv_res;
    }
};