思路:

1. 对于每一个节点, 返回其所在子树所能提供的最大值, 且该最大值必须是单支的, WA 过

2. max(0, max(lf, rt))+root->val, 可以仅返回根节点, WA 过

3. 需要维护一个全局最优解 ans, WA 过

代码:

class Solution {public:	int ans;	int solve_dp(TreeNode *root) {		if(root == NULL)			return 0;		int sum = root->val;		int lf = 0, rt = 0;		if(root->left)		lf = solve_dp(root->left);				if(root->right) 		rt = solve_dp(root->right);		if(lf > 0)			sum += lf;		if(rt > 0)			sum += rt;		ans = max(ans, sum);		return max(0, max(lf, rt))+root->val;	}    int maxPathSum(TreeNode *root) {		ans = -100000000;		solve_dp(root);		return ans;    }};