标签：
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.
求二叉树的最大路径和，感觉好复杂，但是分析一下，由于中间的点不能重叠，所以说肯定一部分在某个点的左边一个在右边。所以可以遍历左右子树来求最大值，相当于遍历所有的单点以及其左右子树之后，不断的更新最大值，用ret全局变量来维护这个最大值。将总体当作根节点加上左边和右边就可以了，代码如下：
1 class Solution { 2 public: 3 int maxPathSum(TreeNode *root) { 4 if(!root) return ret; 5 ret = INT_MIN; 6 dfs(root); 7 return ret; 8 } 9 10 int dfs(TreeNode * node) 11 { 12 if(node == NULL) return 0; 13 int val = node->val; 14 int left = dfs(node->left); 15 int right = dfs(node->right); 16 if(left > 0) val += left; 17 if(right > 0) val += right; 18 if(val > ret) 19 ret = val; 20 return max(node->val, max(node->val + left, node->val + right)); 21 } 22 private: 23 int ret; 24 };
LeetCode OJ：Binary Tree Maximum Path Sum（二叉树最大路径和）
标签：
原文地址：http://www.cnblogs.com/-wang-cheng/p/4918464.html
