Merge pull request #568 from sr-sweta/Contribute

Binary Tree Maximum Path Sum using CPP

Author: smv1999

Data Structures/Trees/BinaryTree_Maximum_Path_Sum.cpp

@@ -0,0 +1,56 @@
+/*
+Src : LeetCode
+--------------
+
+A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence at most once. Note that the path does not need to pass through the root.
+The path sum of a path is the sum of the node's values in the path.
+Given the root of a binary tree, return the maximum path sum of any path.
+
+Input: root = [-10,9,20,null,null,15,7]
+Output: 42
+Explanation: The optimal path is 15 -> 20 -> 7 with a path sum of 15 + 20 + 7 = 42.
+*/
+
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+class Solution {
+public:
+	/**The logic is very simple. The maximum path sum will always consist of a node and its left and right sub-tree maximum path sum. As the node values can be negative also so 
+	4 cases comes here .
+	1) A node and its both sub-trees will constitute the sum 
+	2) Only a node and its left sub-tree will constitute the sum
+	3) Only a node and its right sub-tree will constitute the sum
+	4) Only one node will constitute the sum
+	*/
+    int maxPathSum(TreeNode* root) {
+        if(root==NULL)
+            return 0;
+        int ans=root->val,sum;
+        sum=helper(root,ans);
+        return max(ans,sum);
+    }
+    
+    int helper(TreeNode* root,int &ans){
+        if(root==NULL)
+            return 0;
+        int l=helper(root->left,ans);
+        int r=helper(root->right,ans);
+        if(l<0)
+            l=0;
+        if(r<0)
+            r=0;
+        //Case 1 and 4 is checked here
+        ans=max(ans,l+r+root->val);
+        //Case 2 and 3 is checked here
+        return max(l+root->val,r+root->val);
+    }
+};