solutions: 0404 - Sum of Left Leaves (Easy)

Author: wingkwong

solutions/0400-0499/0404-sum-of-left-leaves-easy.md

@@ -1,5 +1,6 @@
 ---
-description: 'Author: @heiheihang | https://leetcode.com/problems/sum-of-left-leaves/'
+description: 'Author: @heiheihang, @wingkwong, @jit | https://leetcode.com/problems/sum-of-left-leaves/'
+tags: [Tree, Depth-First Search, Breadth-First Search, Binary Tree]
 ---
 
 # 0404 - Sum of Left Leaves (Easy)
@@ -38,27 +39,63 @@ Output: 0
 
 Following the template from [DFS Guide](../../tutorials/graph-theory/depth-first-search), we iterate to each node and check if it is a left leaf. If it is a left leaf, we return that value to its parent for cumulating the sum. The trick here is to add a `is_left` parameter to the `dfs` function.
 
+<Tabs>
+<TabItem value="py" label="Python">
 <SolutionAuthor name="@heiheihang"/>
 
-```python
+```py
 def sumOfLeftLeaves(self, root: Optional[TreeNode]) -> int:
-        def dfs(node, isLeft):
-            
-            #skip when we reach the end of the path
-            if(node is None):
-                return 0
-                
-            #if node is a leaf (no left and right) and is the left child
-            #return its value for adding
-            if(node.left is None and node.right is None and isLeft is True):
-                return node.val 
-            
-            #perform dfs on both children
-            #set isLeft to true on left child
-            #set isLeft to false on right child
-            return dfs(node.left, True) + dfs(node.right, False)
-        
-        #root can be a leaf if it is the only node in the tree
-        #but it is not a left leaf (no right/left)
-        return dfs(root, False)
+    def dfs(node, isLeft):
+        # skip when we reach the end of the path
+        if (node is None): return 0
+        # if node is a leaf (no left and right) and is the left child
+        # return its value for adding
+        if (node.left is None and node.right is None and isLeft is True): return node.val 
+        # perform dfs on both children
+        # set isLeft to true on left child
+        # set isLeft to false on right child
+        return dfs(node.left, True) + dfs(node.right, False)
+    # root can be a leaf if it is the only node in the tree
+    # but it is not a left leaf (no right/left)
+    return dfs(root, False)
 ```
+
+</TabItem>
+
+
+<TabItem value="cpp" label="C++">
+<SolutionAuthor name="@wingkwong"/>
+
+```cpp
+class Solution {
+public:
+    int sumOfLeftLeaves(TreeNode* root, bool left = false) {
+        // use left flag to determine it is a left node
+        // traverse each node and only return if left=true and no child branches
+        if(!root) return 0;
+        if(left && !root->left && !root->right) return root->val;
+        return sumOfLeftLeaves(root->left, true) + sumOfLeftLeaves(root->right, false);
+    }
+};
+```
+
+</TabItem>
+
+
+<TabItem value="elixir" label="Elixir">
+<SolutionAuthor name="@jit"/>
+
+```elixir
+defmodule Solution do
+  # Using a left/right flag:
+  @spec sum_of_left_leaves(root :: TreeNode.t | nil) :: integer
+  def sum_of_left_leaves(root), do: dfs(root, :neither)
+
+  defp dfs(nil, _flag), do: 0
+  defp dfs(%{val: v, left: nil, right: nil}, :left), do: v
+  defp dfs(%{left: l, right: r}, _flag), do:
+    dfs(l, :left) + dfs(r,
+```
+
+</TabItem>
+</Tabs>