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.

思路:这一题同样采用的bottom up的方式。递归调用help函数。如果传入的根节点为空，直接返回0.设置val为左右子树到当前根节点的最大值，递归求左右子树的路径之和，得到左右子树到当前节点的最大路径和，如果值大于result，就更新result。最终返回经过当前节点的最大路径之和，请注意只能加上左右子树的一支，否则就可能会出现重复加的情况。

1. /**
    * 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 {
   private:
       int result;
   public:
       int help(TreeNode *root){
           if(!root)
               return 0;
           int val =root->val;
           int left=help(root->left);
           int right=help(root->right);
           if(left>0)
               val+=left;
           if(right>0)
               val+=right;
           result=max(result,val);
           return max(root->val,max(root->val+left,root->val+right));
       }
       int maxPathSum(TreeNode* root) {
           if(!root)
               return 0;
           result=INT_MIN;
           help(root);
           return result;
       }
   };