leetcode

commit 18a386dc9d35a170751d6113485da81fcf84a2bb
parent 82eae3c6ca946a335ed1b562593d1bbffb3d6bc2
Author: AndrewLockVI <andrewlaack1@gmail.com>
Date:   Sat, 29 Apr 2023 09:44:46 -0500

Completed binary tree level order using BFS in JS

Diffstat:
A
binary-tree-level-order/binary-tree-level-order.js
 | 
40
++++++++++++++++++++++++++++++++++++++++

1 file changed, 40 insertions(+), 0 deletions(-)
diff --git a/binary-tree-level-order/binary-tree-level-order.js b/binary-tree-level-order/binary-tree-level-order.js
@@ -0,0 +1,40 @@
+//Given a binary tree return the values in it as an array of arrays
+//where each array is a single layer of the tree.
+//Ex.    
+//      2
+//     3 4
+//    6   8
+//Output of the above tree would be: [[2][3,4][6,8]]
+//My solution uses BFS to traverse each layer adding all of the values to an array
+//the time complexity of this is O(n) where n is the number of listnodes in the tree.
+//Time: 60ms Beats: 82.33%
+//Memory: 44.8MB Beats: 12.51%
+
+var levelOrder = function(root) {
+    if(root == null){
+        return [];
+    }
+    if(root.left == null && root.right == null){
+        return [[root.val]];
+    }
+    let levels = [];
+    let queue = new Array();
+    let level = 0
+    queue.splice(0,0,root);
+    while(queue.length != 0){
+        let start_len = queue.length;
+        levels.push([]);
+        for(let i = 0 ; i < start_len ; ++i){
+            let curr_node = queue.pop();
+            levels[level].push(curr_node.val);
+            if(curr_node.left != null){
+                queue.splice(0,0,curr_node.left);
+            }
+            if(curr_node.right != null){
+                queue.splice(0,0,curr_node.right);
+            }
+        }
+        level += 1;
+    }
+    return levels;
+};