leetcode

commit e58286d7a6e800ab3a8e535874a79b32243bc9f6
parent 7e8db1c42c55ac40ae3c7e1bbbdd3e417717e8d7
Author: AndrewLockVI <andrewlaack1@gmail.com>
Date:   Fri,  5 May 2023 07:51:41 -0500

Completed find mode of binary tree using dart

Diffstat:
A
find-mode-in-binary-search-tree/mode-of-binary-tree.dart
 | 
47
+++++++++++++++++++++++++++++++++++++++++++++++

1 file changed, 47 insertions(+), 0 deletions(-)
diff --git a/find-mode-in-binary-search-tree/mode-of-binary-tree.dart b/find-mode-in-binary-search-tree/mode-of-binary-tree.dart
@@ -0,0 +1,47 @@
+//Given a binary search tree find the mode value(s) and
+//return it in a list. To solve this problem I iterated through
+//the tree and added each value to a map where the value was the number
+//of instances of a given number. Then I iterated through the map
+//to find the value with the highest number of occurences and returned 
+//the associated key.
+//The time complexity of this code is O(n) where n is the number of nodes.
+//Time: 307ms Beats: 100%
+//Memory: 145.7MB Beats: 100%
+//Not enough submissions for comparsion metrics.
+
+/**
+ * Definition for a binary tree node.
+ * class TreeNode {
+ *   int val;
+ *   TreeNode? left;
+ *   TreeNode? right;
+ *   TreeNode([this.val = 0, this.left, this.right]);
+ * }
+ */
+class Solution {
+  List<int> findMode(TreeNode? root) {
+      int usages = 0;
+      List<int> return_vals = [0];
+      recurse(root);
+      for(var entries in vals.entries){
+          if(entries.value > usages){
+              usages = entries.value;
+              return_vals = [entries.key];
+          }
+          else if(entries.value == usages){
+              return_vals.add(entries.key);
+          }
+      }
+      return return_vals;
+  }
+  Map<int, int> vals = {};
+
+  void recurse(TreeNode? root){
+      if(root == null){
+          return;
+      }
+      vals[root.val] = (vals[root.val] ?? 0) + 1;
+      recurse(root.left);
+      recurse(root.right);
+  }
+}