# Leetcode Binary Tree Maximum Path Sum

Source: Internet
Author: User

Given a binary tree, find the maximum path sum.

The path may start and end at any node in the tree.

For example:
Given the below binary tree,

`       1      /      2   3`

Return `6` .

`1 /**2 * Definition for binary tree3 * public class TreeNode {4 * int val;5 * TreeNode left;6 * TreeNode right;7 * TreeNode (int x) {val = x;}8  * }9  */Ten  Public classSolution { One intMax =Integer.min_value; A      Public intmaxpathsum (TreeNode root) { -         if(Root = =NULL) { -             return0; the         } - maxsum (root); -         returnMax; -     } +  -     Private intmaxsum (TreeNode root) { +         if(Root = =NULL) { A             return0; at         } -         intLeftsum=0; -         intRightsum=0; -         intVal=Root.val; -         if(Root.left! =NULL) { -Leftsum =maxsum (root.left); in             if(Leftsum > 0) { -val = val +leftsum; to             } +         } -  the         if(Root.right! =NULL) { *Rightsum =maxsum (root.right); \$             if(Rightsum > 0) {Panax Notoginsengval = val +rightsum; -             } the         } +  A         if(Val >max) { theMax =Val; +         } -  \$         returnMath.max (Root.val, Math.max (Root.val + leftsum, Root.val +rightsum)); \$  -  -     } the}`

Leetcode Binary Tree Maximum Path Sum

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