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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。
/** * 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; } };
124.Binary Tree Maximum Path Sum
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原文地址：http://www.cnblogs.com/zhoudayang/p/5043379.html
