Merge pull request #7 from FreeTymeKiyan/Kuang

Add Solution for Problem 124

Author: forest5013

C++/124_Binary_Tree_Maximum_Path_Sum.cpp

@@ -0,0 +1,68 @@
+// 124. Binary Tree Maximum Path Sum
+/**
+ * Given a binary tree, find the maximum path sum.
+ *
+ * For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path does not need to go through the root.
+ *
+ * For example:
+ * Given the below binary tree,
+ *
+ *       1
+ *      / \
+ *     2   3
+ * Return 6.
+ *
+ * Tags: Tree, Depth-first Search
+ *
+ * Author: Kuang Qin
+ */
+
+#include "stdafx.h"
+#include <windows.h>
+
+using namespace std;
+
+/**
+ * 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 {
+    int curMax;
+
+public:
+    int maxPathSum(TreeNode* root) {
+        curMax = INT_MIN;
+        dfs(root);
+        return curMax;
+    }
+
+// if we look at each individual nodes, there are only 3 possiblities:
+//   a) Left branch:   root->val + left
+//   b) Right branch:  root->val + right
+//   c) Both branches: root->val + left + right
+//
+// a) or b): it can be accumulated between different nodes, so we use them as the return value in the recursion
+// c):       it can be only a local maximum, so we compare it to the global maximum and keep whichever is larger
+
+    int dfs(TreeNode* root) {
+        if (!root) return 0;
+
+        int left  = max(dfs(root->left), 0);    // if left child val < 0, keep only the parent node value
+        int right = max(dfs(root->right), 0);   // if right child val < 0, keep only the parent node value
+        curMax = max(curMax, root->val + left + right);  // compare the global max to possible local max
+
+        return root->val + max(left, right);    // choose whichever is larger
+    }
+};
+
+int _tmain(int argc, _TCHAR* argv[])
+{
+	return 0;
+}
+