LeetCode OJ：Binary Tree Maximum Path Sum（二叉树最大路径和）

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：

class Solution {
public:
    int maxPathSum(TreeNode *root) {
        if(!root) return ret;
        ret = INT_MIN;
        dfs(root);
        return ret;
    }

    int dfs(TreeNode * node)
    {
        if(node == NULL) return 0;
        int val = node->val;
        int left = dfs(node->left);
        int right = dfs(node->right);
        if(left > 0) val += left;
        if(right > 0) val += right;
        if(val > ret)
            ret = val;
        return max(node->val, max(node->val + left, node->val + right));
    }
private:
    int ret;
};