Leetcode - Binary Tree Maximum Path Sum

解题的关键在于这条路径只能是先往上走，到达某个最高点，再往下走，换句话说，只能有一次转折的机会。所以递归这棵树，记录以某个子节点为转折点时的最大值。值得注意的是树节点的值有负值，所以如果某个子路径的和小于0，放弃它（设置和为0）。

class Solution {
public:
int maxPathSum(TreeNode *root) {

int maxSum = -1 << 30;

int leftMax = pathMaxSum(root->left, maxSum);
if (leftMax < 0)
leftMax = 0;
int rightMax = pathMaxSum(root->right, maxSum);
if (rightMax < 0)
rightMax = 0;

int pathSum = leftMax + rightMax + root->val;
if (pathSum > maxSum)
return pathSum;
else
return maxSum;
}

int pathMaxSum(TreeNode* node, int& maxSum)
{
if (node == NULL)
return 0;

int leftMax = pathMaxSum(node->left, maxSum);
if (leftMax < 0)
leftMax = 0;

int rightMax = pathMaxSum(node->right, maxSum);
if (rightMax < 0)
rightMax = 0;

if (leftMax + rightMax + node->val > maxSum)//turn down at this point
maxSum = leftMax + rightMax + node->val;

int pathMax = leftMax > rightMax ? leftMax : rightMax;
pathMax += node->val;

return pathMax;
}
};