New Problem Solution "Sum of Left Leaves"

Author: haoel

README.md

@@ -8,6 +8,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|404|[Sum of Left Leaves](https://leetcode.com/problems/sum-of-left-leaves/) | [C++](./algorithms/cpp/removeKDigits/sumOfLeftLeaves/SumOfLeftLeaves.cpp)|Easy|
 |402|[Remove K Digits](https://leetcode.com/problems/remove-k-digits/) | [C++](./algorithms/cpp/removeKDigits/RemoveKDigits.cpp)|Medium|
 |401|[Binary Watch](https://leetcode.com/problems/binary-watch/) | [C++](./algorithms/cpp/binaryWatch/BinaryWatch.cpp)|Easy|
 |400|[Nth Digit](https://leetcode.com/problems/nth-digit/) | [C++](./algorithms/cpp/nthDigit/NthDigit.cpp)|Easy|

algorithms/cpp/removeKDigits/sumOfLeftLeaves/SumOfLeftLeaves.cpp

@@ -0,0 +1,71 @@
+// Source : https://leetcode.com/problems/sum-of-left-leaves/
+// Author : Hao Chen
+// Date   : 2016-11-12
+
+/*************************************************************************************** 
+ *
+ * Find the sum of all left leaves in a given binary tree.
+ * 
+ * Example:
+ * 
+ *     3
+ *    / \
+ *   9  20
+ *     /  \
+ *    15   7
+ * 
+ * There are two left leaves in the binary tree, with values 9 and 15 respectively. 
+ * Return 24.
+ ***************************************************************************************/
+
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+ * };
+ */
+class Solution {
+public:
+
+    
+    void sumOfLeftLeaves_recursion_v1(TreeNode* root, int& result) {
+        if (root == NULL ) {
+            return;
+        }
+        
+        if (root->left && root->left->left == NULL && root->left->right == NULL) {
+            result += root->left->val;
+        }
+        sumOfLeftLeaves_recursion_v1(root->left, result);
+        sumOfLeftLeaves_recursion_v1(root->right, result);
+        
+    }
+    
+    int sumOfLeftLeaves_recursion_v2(TreeNode* root) {
+        if (root == NULL ) {
+            return 0;
+        }
+        int result = 0;
+        if (root->left && root->left->left == NULL && root->left->right == NULL) {
+            result = root->left->val;
+        }
+        result += sumOfLeftLeaves_recursion_v2(root->left) + sumOfLeftLeaves_recursion_v2(root->right);
+        return result;
+    }    
+    
+
+    int sumOfLeftLeaves(TreeNode* root) {
+        srand(time(NULL));
+        if (rand()%2) {
+            int result = 0;
+            sumOfLeftLeaves_recursion_v1(root, result);
+            return result;
+        } else {
+            return sumOfLeftLeaves_recursion_v2(root);
+        }
+        
+    }
+};