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binary-tree-level-order.js (1198B)

      1 //Given a binary tree return the values in it as an array of arrays
      2 //where each array is a single layer of the tree.
      3 //Ex.    
      4 //      2
      5 //     3 4
      6 //    6   8
      7 //Output of the above tree would be: [[2][3,4][6,8]]
      8 //My solution uses BFS to traverse each layer adding all of the values to an array
      9 //the time complexity of this is O(n) where n is the number of listnodes in the tree.
     10 //Time: 60ms Beats: 82.33%
     11 //Memory: 44.8MB Beats: 12.51%
     12 
     13 var levelOrder = function(root) {
     14     if(root == null){
     15         return [];
     16     }
     17     if(root.left == null && root.right == null){
     18         return [[root.val]];
     19     }
     20     let levels = [];
     21     let queue = new Array();
     22     let level = 0
     23     queue.splice(0,0,root);
     24     while(queue.length != 0){
     25         let start_len = queue.length;
     26         levels.push([]);
     27         for(let i = 0 ; i < start_len ; ++i){
     28             let curr_node = queue.pop();
     29             levels[level].push(curr_node.val);
     30             if(curr_node.left != null){
     31                 queue.splice(0,0,curr_node.left);
     32             }
     33             if(curr_node.right != null){
     34                 queue.splice(0,0,curr_node.right);
     35             }
     36         }
     37         level += 1;
     38     }
     39     return levels;
     40 };