leetcode

commit 897a722df2a57e8aaac99f88a8eaf41c712022e8
parent e75c5c8414f0254d98d1b7ba205820085e966067
Author: AndrewLockVI <andrewlaack1@gmail.com>
Date:   Sun, 30 Apr 2023 17:40:43 -0500

Completed sum of left leaves using dart

Diffstat:
A
sum-of-left-leaves/sum-of-left-leaves.dart
 | 
38
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1 file changed, 38 insertions(+), 0 deletions(-)
diff --git a/sum-of-left-leaves/sum-of-left-leaves.dart b/sum-of-left-leaves/sum-of-left-leaves.dart
@@ -0,0 +1,38 @@
+//Find the sum of all leaves that are left of their parent.
+//to do this I used DFS to recurse through the list and if the current node is 
+//a leaf it checks if it is left. If it is a left leaf then the stored value of it is
+//added onto the total integer. The time complexity of this code is O(n) where n is the number
+//of nodes in the tree. 
+//Runtime: 266ms Beats: 100%
+//Memory: 142.3MB Beats: 100%
+
+
+/**
+ * Definition for a binary tree node.
+ * class TreeNode {
+ *   int val;
+ *   TreeNode? left;
+ *   TreeNode? right;
+ *   TreeNode([this.val = 0, this.left, this.right]);
+ * }
+ */
+class Solution {
+  int total = 0;
+  int sumOfLeftLeaves(TreeNode? root) {
+      recurse(root ?? new TreeNode(), false);
+      return total;
+  }
+  
+  void recurse(TreeNode node, bool is_left ){
+      if(node.left != null){
+          recurse(node.left ?? new TreeNode(), true);
+      }
+      if(node.right != null){
+          recurse(node.right ?? new TreeNode(), false);
+      }
+      if(is_left && node.right == null && node.left == null){
+          total += node.val;
+      }
+  }
+
+}